sin .cos ∫ - mt educare,mteducaresouth.com/images/maths/homework/indefinite-integration-13... ·...
TRANSCRIPT
XI – CLASS WORK PROBLEMS
Q-1) (((( ))))8
79 – + 4 – 3
2
∫∫∫∫
xx dx
Ans. Let I = ( )8
79 – + 4 – 32
xx dx
∫
=( )
××××
9 –4 – 3 12
+ +1 8 4
9 –2
x
xc
9
8
=( )
4 – 32– 9 – + +9 2 32
xxc
89
Q-2)2sin 3∫∫∫∫ x dx
Ans. I =2sin 3x dx∫
=1 – cos6
2
xdx∫
=1
– cos62
dx x dx ∫ ∫
=1 sin6
– +2 6
xx c
=sin6– +
2 12
x xc
Q-4) sin
1+ sin∫∫∫∫x
dxx
Ans. Let I =sin
1+sin
xdx
x∫=
( )
( ) ( )
sin 1 – sin
1+sin 1 – sin
x xdx
x x∫
=( )
2
sin 1 – sin
1 – sin
x xdx
x∫=
( )2
sin 1 – sin
cos
x xdx
x∫= ( )2tan .sec – tanx x x dx∫= ( )2tan .sec – sec –1x x x dx
∫= ( )2tan .sec – sec +1x x x dx∫= secx – tanx + x + c.
Q-3) (((( ))))2
tan3 cot3++++∫∫∫∫ x x dx
Ans. I = ( )2
tan3 + cot3x x dx∫= ( )2 2tan 3 +2tan3 cot3 + cot 3x x x x dx∫= ( )2 2sec 3 –1+2+ cos 3 –1x ec x dx∫= ( )2 2sec 3 + cos 3x ec x dx∫=
2 2sec 3 + cosec 3x x dx∫ ∫Indefinite Integration
= tan3 cot 3
– +3 3
x xc
Q-5) 2 2
1
sin .cos∫∫∫∫ dxx x
Ans. Let I = 2 2
1
sin .cosdx
x x∫=
2 2
2 2
sin + cos
sin .cos
x xdx
x x∫
=
2 2
2 2 2 2
sin cos+
sin .cos sin .cos
x xdx
x x x x
∫
= 2 2
1 1+
cos sindx
x x
∫
= ( )2 2sec + cosx ec x dx∫= tanx – cotx + c
Indefinite Integration
2 Mahesh Tutorials Science
Q-8) If f ′′′′(x) = 4x3 – 2x + 1 and f(2) = 17, find f(x)
Ans. = ( )'f x dx∫= ( )34 – 2 +1x x dx∫=4 2
– + +4 2
x xx c
4 2
Q-9)
26 +17 +10
2 +3∫∫∫∫x x
dxx
Ans. Let I =
26 +17 +10
2 + 3
x xdx
x∫
2
2 ±±±±
±±±±
3 + 4
2 +3 6 +17 +10
–6 9
8 10
–8 12
–2
x
x x x
x x
x
x
+
∴∴∴∴ I = ( )( )
( )
–23 + 4 +
2 + 3x dx
x
∫
=
2
××××3 1
+ 4 – 2log 2 +3 +2 2
xx x c
=
23+ 4 – log 2 + 3 +
2
xx x c
= x2 + 3x +7 log |7x – 1|+ c
Q-10)6 – 8
5 –1∫∫∫∫x
dxx
Ans. I = 6 – 8
5 –1
xdx
x∫Let 6x – 8 = a (5x – 1) + b
6x – 8 = 5ax – (a – b)
Comparing co-efficients
5a = 6 a – b = 8
a = 6
5
6
5– b = 8 ∴∴∴∴ b =
–34
5
I = ( )
( )
6 345 –1 –
5 5
5 –1
x
dxx∫
= ( )
6 34–
5 5 5 –1dx
x
∫
=6 34 1
–5 5 5 –1
dx dxx∫ ∫
=××××
6 34– log 5 –1 +
5 5 5x x c
=6 34– log 5 –1 +
5 25
xx c
Q-6) cos – 4 tan
4 cos∫∫∫∫x x
x
e x xdx
x
Ans. Let I =cos – 4 tan
4 cos
e x xdx
x∫x x
x
=cos 4 tan
–4 cos 4 cos
e x xdx
x x
∫
x x
x x
= – tan .sec4
ex x dx
∫
x
=4
– sec +
log4
e
x ce
x
=( )
– sec +4 log – log 4
ex c
e
x
x
=( )
– sec +4 1 – log 4
ex c
x
x
Q-7)cos2 – cos2
cos – cos
αααα
αααα∫∫∫∫x
dxx
Ans. =( ) ( )2 2 αααα
αααα
2cos –1 – 2cos –1
cos – cos
xdx
x∫=
2 2αααα
αααα
2cos –1 – 2cos +1
cos – cos
xdx
x∫
=( )2 2 αααα
αααα
2 cos – cos
cos – cos
xdx
x∫=
( ) ( )α αα αα αα α
αααα
cos – cos cos + cos2
cos – cos
x xdx
x∫= ( )αααα2 cos + cosx dx∫= αααα2 cos + cos 1x dx dx∫ ∫= 2.sin x + 2 cos αααα × × × × x + c
= 2 sin x + 2x cos α α α α + c
∴∴∴∴ f(x) = x4 – x2 + x + c
∴∴∴∴ f(x) = (2)4 – (2)2 + 2 + c
= 16 – 4 + 2 + c
∴∴∴∴ f(x) = x4 – x2 + x + 3
Mahesh Tutorials Science 3
Indefinite Integration
XI – HOME WORK PROBLEMS
Q-1)
3 2+ 3 + +2
∫∫∫∫x x x
dxx
Ans. Let I =
3 2+3 + +2x x xdx
x∫=
2 2+3 +1+x x dx
x
∫
=3+ + +2log +
3 2
x xx x c
3 2
Q-2)
3 +2 + 5
∫∫∫∫x x
dxx
Ans. Let I =
3 +2 +5x xdx
x∫
=
2 2 5+ +
x xdx
x x x
∫
=
3 1 –1
2 2 2+ 2 +5x x x dx ∫
=
5 3 1
2 2 22 5+ + +
5 3 1
2 2 2
x x xc
=
5 3 1
2 2 22 4
+ +10 +5 3x x x c
Q-3)1
3 – 4∫∫∫∫ dxx
Ans. Let I =1
3 – 4dx
x∫
= ( )
–1
2
3 – 4x dx∫
=
( )
1
2
××××3 – 4 1
+1 –4
2
xc
= ( )
1
21– 3 – 4 +2
x c
Q-4)3cos∫∫∫∫ x dx
Ans. Let I =3cos x dx∫
=cos3 +3cos
4
x xdx∫
=1
cos3 + 3cos4
x dx x dx ∫ ∫
= ( )1 sin3
+ 3 sin +4 3
xx c
∴∴∴∴ I =1 sin3
+3sin +4 3
xx c
Q-5) (((( ))))
3
2
6 +5∫∫∫∫ x dx
Ans. Let I = ( )
3
2
6 +5x dx∫
=
( )
5
2
××××6 +5 1
+5 6
2
xc
= ( )
5
216 +5 +
15x c∫
Q-6) (((( )))) (((( ))))4 7
3 +5 + 5 – 3 ∫∫∫∫ x x dx
Ans. Let I = ( ) ( )4 7
3 +5 + 5 – 3x x dx ∫
=( ) ( )
5 8
× ×× ×× ×× ×3 +5 5 – 31 1
+ +5 3 8 –3
x xc
= ( ) ( )5 81 1
3 +5 – 5 – 3 +15 24
x x c
Q-7)+ 4 –∫∫∫∫dx
x x
Ans. Let I =+ 4 –
dx
x x∫
= ( )( )( )
××××+ 4 +
+ 4 – + 4 +
x xdx
x x x x∫
= ( )
+ 4 –
+ 4 –
x xdx
x x∫
Indefinite Integration
4 Mahesh Tutorials Science
Q-12)1
1 – cosdx
x∫∫∫∫Ans. Let I =
1
1 – cosdx
x∫=
( )
( ) ( )
1+ cos
1 – cos 1+ cos
xdx
x x∫
=( )
2
1+ cos
1 – cos
xdx
x∫= 2
1+ cos
sin
xdx
x∫= ( )2cosec + cot .cosecx x x dx∫= –cot x – cosec x + c
Q-13)cos2
1+ cos2∫∫∫∫x
dxx
Ans. Let I =cos2
1+ cos2
xdx
x∫=
2
2
2cos –1
2cos
xdx
x∫= 2
11 –2cos
dxx
∫
=21
1 – sec2
dx x dx∫ ∫= x –
1
2 tan x + c
Q-9)2cos∫∫∫∫ x xdx
Ans. Let I =2cos x x dx∫
=1+ cos2
2
xdx∫
=1
1 + cos22
dx x dx ∫ ∫
= 1 sin2
+ +4 2
xx c
=1 sin2
+ +2 2
xx c
Q-10) sin5 .cos3∫∫∫∫ x xdx
Ans. Let I = sin5 .cos3x x dx∫=
12sin5 .cos3
2x x dx∫
= ( ) ( )1
sin 5 + 3 +sin 5 – 32
x x x x dx ∫= ( )1
sin8 +sin24
x x dx∫=1 – cos8 cos2
– +2 8 2
x xc
Q-11) 2
4 – 7cos3
sin 3
xdx
x∫∫∫∫Ans. Let I = 2
4 – 7cos3
sin 3
xdx
x∫= 2 2
4 7cos3–
sin 3 sin 3
xdx
x x
∫
=24 cosec 3 – 7x dx∫
cot3 .cosec3x x dx∫=
( )– cos 34cot3– – 7 +
3 3
ec xxc
=4 7– cot 3 + cosec3 +3 3
x x c
= ( )1
+4 +4
x x dx∫
=( )
33
22+ 41
+ +3 34
2 2
x xc
= ( )
33
22
1+ 4 + +
6x x c
Q-8)2cot∫∫∫∫ xdx
Ans. Let I =2cot x dx∫
= ( )2cosec –1x dx∫= – cot x – x + c
Mahesh Tutorials Science 5
Indefinite Integration
Q-14) –1 cos2
tan1+ sin
xdx
x
∫∫∫∫
Ans. Let I =–1 cos2
tan1+sin
xdx
x
∫
=
2 2
–1
2
cos – sin2 2tan
cos + sin2 2
x x
dxx x
∫
=–1
cos – sin2 2tan
cos + sin2 2
x x
dxx x
∫
=–1
1 – tan2tan
1+ tan2
x
dxx
∫
= ( )–1 –1tan 1 – tan tan2
xdx
∫
=ππππ–4 2
xdx
∫
=2ππππ
– +4 4
xx c
Q-15)If f′′′′(x) = 4x3 – 3x2 + 2x + k and f (0) = 1,
f (1) = 4, find f(x).
Ans. By the definition of integral,
f′(x) = ( ) ( )' = 4 – 3 + 2 +f x dx x x x k dx∫ ∫ 3 2
= 4 – 3 + 2 + 1x dx x dx x dx k dx∫ ∫ ∫ ∫3 2
= 4 – 3 + 2 + +4 3 2
x x xkx c
4 3 2
∴∴∴∴ f(x) = x4 – x3 + x2 + kx + c ... (i)
Now f(0) = 1 gives
f(0) = 0 + 0 + 0 + 0 + c = 1
∴∴∴∴ c = 1
∴∴∴∴ from (i), f(x) = x4 – x3 + x2 + kx + 1 ... (ii)
Further f(1) = 4 gives
f(1) = 1 – 1 + 1 + k + 1 = 4
∴∴∴∴ k = 2
∴∴∴∴ from (ii), f(x) = x4 – x3 + x2 + 2x + 1.
Q-16)sin + 5 cot
5 sin∫∫∫∫x x
x
e x xdx
x
Ans. Let I =sin +5 cot
5 sin
e x xdx
x∫x x
x
= + cot .cosec5
ex x dx
∫
x
x
= + cot .cosec5
edx x x dx
∫ ∫
x
=
5– cosec +
log5
e
x ce
∫
x
= ( )– cosec +
5 log – log5
ex c
e∫x
x
= ( )– cosec +
5 1 – log5
ex c∫
x
x
Q-17)
4 2– 3 + 4
∫∫∫∫x x x
dxx
Ans. Let I =
4 2– 3 + 4x x xdx
x∫
=
4 23 4– +
x x xdx
x x x
∫
= ( )3 – 3 + 4x x dx∫=
4 23– + 4 +
4 2
x xx c
Q-18) (((( )))) (((( ))))2 + 4 – 3∫∫∫∫ x x dx
Ans. Let I = ( ) ( )2 + 4 – 3x x dx∫= ( )3 2– 3 + 4 –12x x x dx∫=
4 3 23 4– + –12 +
4 3 2
x x xx c
=4
3 2– + 2 –12 +4
xx x x c
Indefinite Integration
6 Mahesh Tutorials Science
Q-24)2tan∫∫∫∫ x dx
Ans. Let I =2tan x dx∫ = ( )2sec –1x dx∫
=2sec –dx dx∫ ∫ = tan x – x + c
Q-25) (((( ))))sin4 – cos6∫∫∫∫ x x dx
Ans. Let I = ( )sin4 – cos6x x dx∫= sin4 – cos6x dx x dx∫ ∫=
cos4 sin6– – +
4 6
x xc
Q-20)
22 +3 – 4
∫∫∫∫x x
dxx
Ans. Let I =
22 +3 – 4x xdx
x∫
∴∴∴∴ I =
22 3 4+ –
x x xdx
x x x
∫
=
3 1 –1
2 2 22 +3 – 4x x x dx ∫
=
5 3 1
2 2 2
2 +3 – 4 +5 3 1
2 2 2
x x xc
=
5 3 1
2 2 24
+2 – 8 +5x x x c
Q-21) (((( ))))5
3 + 7∫∫∫∫ x dx
Ans. Let I = ( )5
3 +7x dx∫=
( )6
××××3 + 7 1
6 3
x
=( )
63 +7
+18
xc∫
Q-22)1
3 – 2∫∫∫∫ dxx
Ans. Let I =1
3 – 2dx
x∫∴∴∴∴ I = ( )
–1
23 – 2x dx∫
Q-23)1
2 +5 – 2 – 3∫∫∫∫ dxx x
Ans. Let I =1
2 +5 – 2 – 3dx
x x∫
= ( ) ( )2 +5 + 2 – 3
2 +5 – 2 – 3 2 +5 + 2 – 3
x xdx
x x x x∫
=( )( ) ( )
2 +5 – 2 – 3
2 +5 – 2 – 3
x xdx
x x∫
=1
2 +5 + 2 – 38
x x dx ∫
= ( ) ( )
3 3
2 2
× ×× ×× ×× ×
2 +5 2 – 31+ +
3 382 2
2 2
x xc
= ( ) ( )
3 3
2 2
2 +5 2 – 31+ +
8 3 3
x xc
= ( ) ( )
3 3
2 212 +5 + 2 – 3 +
24x x c
Q-19) (((( ))))
6 414 – 5 + 7
– 2cot .cos + 5
∫∫∫∫
x x xdx
x e x
Ans. Let I =( )
6 4– +
+
14 5 7
– 2cot .cosec 5
x x xdx
x x x
∫
=
3
7 5 2
14 – 5 + 737 5
2
x x x
( )–2 –cosec +5 +x x c
= 2x7 – x5 +
3
214
3x + 2 cosec x + 5x + c
= ( )1
2
××××3 – 2 1
+1 –2
2
xc
= ( )1
23 – 2 +x c
Mahesh Tutorials Science 7
Indefinite Integration
Q-26)3sin∫∫∫∫ xdx
Ans. Let I =3sin x dx∫
∵∵∵∵ sin 3x = 3sinx – 4sin3x
⇒⇒⇒⇒ sin3x = 3sin x –– sin3
4
x
=1
4[3sinx – sin 3x]
∴∴∴∴ I =1
4[3sinx – sin 3x] dx
=
1 cos3–3cos + +
4 3
xx c
∴∴∴∴ I =3 cos3– cos + +4 12
xx c
Q-27) 2
3+ 4 sin5
cos 5∫∫∫∫xdx
x
Ans. Let I = 2
3+ 4sin5
cos 5
xdx
x∫= 2 2
3 4sin5+
cos 5 cos 5
xdx dx
x x∫ ∫=
23 sec 5 +4 tan5 .sec5x dx x x dx∫ ∫=
tan5 4sec53 + +
5 5
x xc
=1
5(3 tan 5x + 4sec 5x) + c
Q-28) 1+ sin∫∫∫∫ xdx
Ans. Let I = 1+sinxdx∫
=
2
cos +sin2 2
x xdx
∫
=
sin – cos2 2+ +
1 1
2 2
x x
c
= 2sin – 2cos +2 2
x xc
= 2 sin – cos +2 2
x xc
Q-29)
3 23 + 7 + 7 +1
3 +4∫∫∫∫x x x
dxx
Ans. Let I =
3 23 + 7 +7 +1
3 + 4
x x xdx
x∫
=
2
3 2
3 2
2
2
±±±±
±±±±
±±±±
+ +1
3 + 4 3 +7 +7 +1
–3 4
3 +7 +1
– 3 4
3 +1
– 3 4
– 3
x x
x x x x
x x
x x
x x
x
x
∴∴∴∴ I = ( )2 –3+ +1 +
3 + 4x x dx
x
∫
= ( )2 1+ +1 – 3
3 +4x x dx dx
x∫ ∫=
3 2 3log 3 4+ + – +
3 2 3
xx xx c
+
=
3 2
+ + – 3log 3 + 4 +3 2
x xx x c
Q-30)
28 +10 – 2
2 +1∫∫∫∫x x
dxx
Ans. I =
28 +10 – 2
2 +1
x xdx
x∫
2
2 2
±±±±
4 +3
2 +1 8 +10 – 2
8 + 4
6 – 2
6 3
– 5
x
x x x
x x
x
x
I = ( ) ( )
( )
2 +1 4 +3 – 5
2 +1
x xdx
x∫
= ( )
54 +3 –
2 +1x dx
x
∫
=
24 5+3 – log 2 +1 +
2 2
xx x c
I =2 5
2 +3 – log 2 +1 +2
x x x c
Indefinite Integration
8 Mahesh Tutorials Science
GROUP (A) – CLASS WORK PROBLEMS
Q-1)(((( ))))sin log x
dxx∫∫∫∫
Ans. Let I =( )sin log x
dxx∫
Put log x = t
∴∴∴∴1dx
x = dt
∴∴∴∴ I = sin t dt∫= – cos t + c
= – cos (log x) + c
Q-2)sin cosxe x dx∫∫∫∫
Ans. Let I = cose x dx∫ sin x
Put sin x = t
∴∴∴∴ cos x dx = dt
∴∴∴∴ I = e dt∫ t
= et + c
∴∴∴∴ I = esin x + c
Q-3) 2
sin
1+ cos
xdx
x∫∫∫∫
Ans. Let I =sin
1+ cos
xdx
x∫ 2
Put, cos x = t – sinx dx = dt
∴∴∴∴ I =–
1+
dt
t∫ 2
= – tan–1 (t) + c
= – tan–1 (cos x) + c
Q-4)3sin .cosx x dx∫∫∫∫
Ans. Let I = sin .cosx x dx∫ 3
Put sin x = t
∴∴∴∴ cos dx = dt
∴∴∴∴ I = t dt∫ 3
= +4
tc
4
=( )sin
+4
xc
4
=1sin +
4x c4
Q-5)4sec .tanx x dx∫∫∫∫
Ans. Let I = sec .tanx x dx∫ 4
Put tan x = t
∴∴∴∴ sec2x dx = dt
∴∴∴∴ I = ( )sec .tan secx x x dx∫ 2 2
= ( ) ( )1+ tan .tan . secx x x dx∫ 2 2
= ( )1+ .t t dt∫ 2
= ( )+t t dt∫ 3
= + +2 4
t tc
2 4
=tan tan
+ +2 4
x xc
2 4
or
I = ( )sec . sec .tanx x x dx∫ 3
Put sec x = t
∴∴∴∴ sec x. tan x dx = dt
∴∴∴∴ I = t dt∫ 3
∴∴∴∴ I = +4
tc
4
∴∴∴∴ I =1
4sec4 x + c
Q-6)(((( ))))–1 2
4
sin
1+
x x
dx
x
∫∫∫∫
Ans. Let I =( )sin
1+
x xdx
x∫–1 2
4
Put sin–1(x2) = t
∴∴∴∴
( )( )
12
1 –
x dx
x2
2
= dt
∴∴∴∴1 –
x dx
x4 =2
dt
Mahesh Tutorials Science 9
Indefinite Integration
∴∴∴∴ I =2
dtt ∫
=1
+2 2
tc
2
∴∴∴∴ I = +4
tc
2
∴∴∴∴ I =( )( )sin
+4
xc
–1 2
Q-7)(((( ))))
(((( ))))
+
+
log 2 – log
2
x xdx
x x∫∫∫∫
Ans. LetI =( )
( )
log + 2 – log
+ 2
x xdx
x x∫Put log(x + 2) – log x = t
∴∴∴∴1 1
–+2
dxx x
= dt
∴∴∴∴ ( )
– – 2
+ 2
x xdx
x x = dt
∴∴∴∴ ( )+ 2
dx
x x =–2
dt
∴∴∴∴ I = –2
dtt ∫ =
1–2
t dt∫
=1
– +2 2
tc
2
=1
–4[log (x + 2) – log x]2 + c
=1
–4[log x – log (x + 2)]2 + c
=1
– log +4 + 2
xc
x
2
Q-8)(((( ))))
(((( ))))2
1
sin∫∫∫∫x
x
e xdx
xe
+
Ans. Let I = ( )
( )2
x
x
1+
sin
e xdx
xe∫Put xex = t
∴ xex + ex = dt
dx
ex (x + 1)dx = dt
Q-9) (((( ))))
1
log log logdx
x x x∫∫∫∫
Ans. Let I = ( )
1
log log logdx
x x x∫
I =
( )
( )
1
log
log log
dxx x
x∫
∴ I =( )( )
( )
log log
log log
dx
dx dxx∫
= log |log(log x| + c
Q-10)
–1 –1+
+
e x
e x
x edx
x e∫∫∫∫
Ans. Let I =
–1 –1e e
e x
+
+
x edx
x e∫Put xe + ex = t
∴ (exe – 1 + ex ) dx = dt
∴ e (xe – 1 + ex – 1 ) dx = dt
e
∴ I =
dt
e
t
=1
elog |t| + c
=1
elog |xe + ex| + c
∴ I = 2
1
sindtt∫
=2cosec t dt∫
= –cot t + c
I = –cot (xex) + c
Indefinite Integration
10 Mahesh Tutorials Science
Q-12) (((( ))))
2 2
sin2 1
sin cos∫∫∫∫x x
dxa x b x
+
+
Ans. Let I =( )
2 2
sin2 1+
sin + cos
x xdx
a x b x∫Put asin2x + bcos2x = t
differentiate w.r.t.x
2a sinx cosx – 2b cosx sinx = dt
dx
a sin2x – b sin2x = dt
dx
(a – b) sin2x = dt
dx
sin2x dx = ( )–
dt
a b
I = ( )
1
–
dt
t a b∫
= ( )
1
–a blog +t c
I = 1
–a b2 2log sin + cos +a x b x c
Q-13)
2
2
–1
1∫∫∫∫x
x
edx
e +
Ans. Let I =
2
2
x
x
–1
+1
edx
e∫
I =
2
2
x
x
x
x
–1
+1
e
e dxe
e
∫
=
–
–
x x
x x
–
+
e edx
e e∫
I = ( )–
–
x x
x x
+
+
de e
dx dxe e∫
∴∴∴∴ I = –x xlog + +e e c
Q-11) –+∫∫∫∫ n
dx
x x
Ans. Let I = –n
n
1=
1++
dxdx
x xx
x
∫ ∫
= +1
n
n +1
x dx
x∫Put xn+1 + 1 = t
∴∴∴∴ (n + 1)xn dx = dt
∴∴∴∴ xn dx = +1
dt
n
∴∴∴∴ I = ( )
1=
+1 +1
dt dt
n t n t∫ ∫
=1
log ++1
t cn
=+1n1
log +1 + .+1
x cn
GROUP (A) – HOME WORK PROBLEMS
Q-1) 2 1
2 . log 2
2
x
x
xdx
x++++
++++
++++∫∫∫∫Ans. Let I = ∫
+2 .log2
+2
xdx
x2 1
x
x ++++
Put x2 + 2x + 1 = t
(2x + 2x + 1 log 2) dx = dt
(x + 2x log 2) dx = 2
dt
∴∴∴∴ I =1
2∫ dtt
∴∴∴∴ I =1
2 ∫ t dt
=1
2 log |t| + c
∴∴∴∴ I = + +1log 2
2x c2 1x ++++
Q-2) cos x
dxx∫∫∫∫
Ans. Let I = cos x
dxx∫
Put x = t
∴∴∴∴ 1dx
x = 2dt
Mahesh Tutorials Science 11
Indefinite Integration
Q-6)(((( ))))
(((( ))))2
+1
cos
x
x
x edx
xe∫∫∫∫
Ans. Let I =( )
( )+1
cos
x edx
xe∫ 2
x
x
Put xex = t
∴∴∴∴ (xex + ex)dx = dt
i.e, (x + 1)ex dx = dt
∴∴∴∴ I =cos
dt
t∫ 2
= sec t dt∫ 2
= tan t + c
= tan(xex) + c
Q-7)
2
61+
xdx
x∫∫∫∫
Ans. Let I =1+
xdx
x∫2
6
=( )1+
xdx
x∫
2
23
Put x3 = t
∴∴∴∴ x2 dx = 3
dt
∴∴∴∴ I =1
31+
dt
t
∫ 2
=1
3 1+
dt
t∫ 2
=1
3tan–1 (t) + c
∴∴∴∴ I =1
3tan–1 (x3) + c
Q-3) (((( ))))4 5cos 2 +x x dx∫∫∫∫Ans. Let I = ( )∫ cos 2+x x dx4 5
Put 2 + x5 = t
∴∴∴∴ 5x4 dx = dt
∴∴∴∴ x4 dx = 5
dt
∴∴∴∴ I = cos5
dtt ∫
=1sin +
5t c
= ( )1sin 2+ +
5x c5
Q-4)1
–
2
11+
xxe dx
x
∫∫∫∫
Ans. Let I =1
1+e dxx
∫
1–
2
xx
Put x – 1
x = t
∴∴∴∴ 1 – 1
–x
2=
dt
dx
∴∴∴∴1
1+ dxx
2= dt
∴∴∴∴ I = e dt∫ t
= et + c
= +e c
1–xx
Q-5)(((( ))))2 –1 3
6
tan
1+
x xdx
x∫∫∫∫
Ans. Let I =( )tan
1+
x xdx
x∫2 –1 3
6
Put tan–1(x3) = t
∴ I = ( )cos 2t dt∫= 2 cos t dt∫= 2 sin t + c
= 2 sin x + c
∴∴∴∴( )
( )1
31+
x dxx
2
23
= dt
∴∴∴∴1+
xdx
x
2
6 =3
dt
∴∴∴∴ I =3
dtt ∫
=1
+3 2
tc
2
∴∴∴∴ I =1
6 [tan–1 (x3)]2 + c
Indefinite Integration
12 Mahesh Tutorials Science
Q-8)
2–1
2
2tan
1
++++++++
++++ ∫∫∫∫ xxax dx
x
Ans. I =
2–1
2
2tan
1
++
+ ∫x
ax dxx
x
Put x + tan–1x = t
11+
+1dx
x2 = dt
+ 2
+1
xdx
x
2
2 = dt
∴∴∴∴ I = ∫a dtt
=log
a
a
t
+ c
∴∴∴∴ I = +log
ac
a
–1tanx x++++
Q-9) 2
cosec .cot
1+ cosec
x xdx
x∫∫∫∫
Ans. Let I =cosec .cot
1+ cosec
x xdx
x∫ 2
Put cosec x = t
∴∴∴∴ – cosec x. cot x dx = dt
∴∴∴∴ cosec x. cot x dx = –dt
∴∴∴∴ I =–
1+
dt
t∫ 2
= –tan–1 (t) + c
= –tan–1 (cosec x) + c
Q-10) 2
+1
2 + 2 +
xdx
x x∫∫∫∫
Ans. Let I = +1
2+2 +
xdx
x x∫ 2
Put 2 + 2x + x2 = t
∴∴∴∴ (2 + 2x) dx = dt
∴∴∴∴ (x + 1) dx = 2
dt
∴∴∴∴ I =1
2
dt
t
∫
=1
2t dt∫–1
2
=1
+12
2
tc
–1
2
Q-11)(((( ))))
72
1+ sin2
+ sin
xdx
x x∫∫∫∫
Ans. Let I = ( )
7
1+sin2
sin+∫ 2
xdx
x x
Put x + sin2x = t
(1 + 2sinx cos x) dx = dt
(1 + sin 2x) dx = dt
I = ∫1dt
t7
= ∫ t dt– 7
= ∫ +–7 +1
tc
– 7+1
= ∫ +–6
tc
– 6
I =( )+sin
+–6
x xc
–62
= +t c1
2
= 2+2 + +x x c2
Q-12)1
++++∫∫∫∫ dxx x
Ans. Let I =1
+dx
x x∫=
( )∫1
+1dx
x x
I =
( )∫2
2 +1dx
x x
f (x) = ( )+1x
f ′′′′(x) =1
2 x
∴∴∴∴ I = 2log +1 +x c
Mahesh Tutorials Science 13
Indefinite Integration
Q-19)(((( ))))–1
2
tan
1+
xdx
x∫∫∫∫
Ans. Let I =( )tan
1+
xdx
x∫–1
2
Put tan–1 x = t
∴∴∴∴1
1+ x2 dx = dt
∴∴∴∴ I = t dt∫ 2
= +3
tc
3
=( )tan
+3
xc
3–1
Q-20)–1 2
1
sin . 1 –dx
x x∫∫∫∫
Ans. Let I =1
sin . 1 –dx
x x∫ –1 2
Put sin–1 x = t
∴∴∴∴1
1 – x2 dx = dt
∴∴∴∴ I =dt
t∫= log|t| + c
= log|sin–1x| + c
Q-13)2+5
x
x
e
e∫∫∫∫Ans. Let f (x) = 2 + 5ex
f′ (x) = 5ex
∴∴∴∴ I =1 5
5 2+5
e
e∫x
x dx
I =1
5 log |2 + 5ex| + c
Q-14)cos – sin
cos + sin
x xdx
x x∫∫∫∫
Ans. Let I =cos – sin
cos + sin
x xdx
x x∫Let f ′(x) = cos x + sin x
∴∴∴∴ f ′(x) = –sin x + cos x = cos x – sin x
∴∴∴∴ I = log |cos x + sin x| + c
Q-15) 2 2
sin2
4 sin + cos
xdx
x x∫∫∫∫
Ans. Let I =sin2
4sin + cos
xdx
x x∫ 2 2
Let f ′(x) = 4 sin2 x + cos2 x
∴∴∴∴ f ′(x) = 8 sin x cos x –2 cos x. sin x
= 4 sin 2x – sin x
= 3 sin 2 x
∴∴∴∴ I =1 3sin2
3 4sin + cos
xdx
x x∫ 2 2
=1
3log|4sin2x + cos2x| + c
Q-17)5sin .cosx x dx∫∫∫∫
Ans. Let I = sin .cosx x dx∫ 5
Put sin x = t
∴∴∴∴ cos x dx = dt
Q-16)cos
sin
xdx
x∫∫∫∫Ans. Let I =
Put sin x = t
∴∴∴∴ cos x dx = dt
∴∴∴∴ I =dt
t∫ = 2 +t c
∴∴∴∴ I = 2 sin +x c
Q-18)5sec .tanx x dx∫∫∫∫
Ans. Let I = sec .tanx x dx∫ 5
= ( )sec . sec tanx x x dx∫ 4
Put sec x = t
∴∴∴∴ sec x. tan x dx = dt
∴∴∴∴ I = t dt∫ 4= +
5
tc
5
∴∴∴∴ I =( )sec
+5
xc
5
I =5sec
+5
xc
∴∴∴∴ I = t dt∫ 5= +
6
tc
6
=( )sin
+6
xc
6
∴∴∴∴ I =( )sin
+6
xc
6
Indefinite Integration
14 Mahesh Tutorials Science
Q-21)2 2+
xdx
x a∫∫∫∫
Ans. Let I =+
xdx
x a∫ 2 2
Put x2 + a2 = t
∴∴∴∴ x dx =2
dt
∴∴∴∴ I =1
2
dt
t
∫ = ( )
12 +
2t c
= +t c = + +x a c2 2
Q-22)
–
–
–
++++∫∫∫∫x x
x x
e e
e e
Ans. I =
–
–
–
+∫e e
e e
x x
x x
f(x) = ex + e–x
f ′′′′(x) = ex – e–x
∴∴∴∴ I = log + +e e c–x x
Q-23) 21
x
x++++∫∫∫∫Ans. I = ∫1+
xdx
x2
I =( )∫2
2 1+
xdx
x2
f (x) = (1 + x2)
f ′′′′(x) = 2 x
∴∴∴∴1
2 log |1 + x2| + c
Q-24)4 2tan .sec .x x dx∫∫∫∫
Ans. Put tan x = t
sec2x dx = dt
∴∴∴∴ I = ∫ t dt4
= +5
tc
5
I =tan
+5
xc
5
Q-25)(((( ))))
3–1
2
sin
1 –
xdx
x∫∫∫∫
Ans. I =( )∫
sin
1 –
xdx
x
3–1
2
Put sin–1 x = t
1
1 –dx
x2 = dt
∴∴∴∴ I = ∫ t dt2
= +4
tc
4
I =1
4 (sin–1 x)4 + c
GROUP (B) – CLASS WORK PROBLEMS
Q-1) (((( ))))
sin
sin –
xdx
x a∫∫∫∫
Ans. Let I = ( )
sin
sin –
xdx
x a∫
=( )
( )
sin – +
sin –
x a adx
x a∫
=( )
( )
sin – +
sin –
x a adx
x a
∫
=
( )
( )
( )
sin – .cos
+cos – .sin
sin –
x a a
x a adx
x a
∫
=( )
( )
sin –cos .
sin –
x aa dx
x a∫
( )
( )
cos –+sin .
sin –
x aa dx
x a∫
= ( )cos . + sin cot –a dx a x a dx∫ ∫= x cos a + sin a log |si n (x – a) + c
Q-2) (((( ))))
(((( ))))
cos +
cos –
x adx
x a∫∫∫∫
Ans. Let I =( )
( )
cos +
cos –
x adx
x a∫
=( )
( )
cos – + 2
cos –
x a adx
x a
∫
Mahesh Tutorials Science 15
Indefinite Integration
GROUP (B) – HOME WORK PROBLEMS
Q-1) (((( ))))
cos
sin +
xdx
x a∫∫∫∫
Ans. Let I = ( )cos
sin +
xdx
x a∫
∴∴∴∴ I =( )( )
( )
cos + –
sin +
x a adx
x a∫∴∴∴∴ I =
( ) ( )
( )
cos + cos +sin + sin
sin +
x a a x a adx
x a∫
∴∴∴∴ I = ( )cos cot + + sin 1a x a dx a dx∫ ∫∴∴∴∴ I = cos a log|sin(x + a)|+ sin a. x + c
∴∴∴∴ I = x .sin a + cos a. log|sin(x + a)|+ c
Q-2)(((( ))))
(((( ))))
sin –
sin ++++∫∫∫∫x a
dxx a
Ans. I =( )
( )
sin –
sin +∫x a
dxx a
=( )
( )
sin – 2
sin
+ +∫
x a adx
x a
=( ) ( )
( )
sin cos2 – cos sin2
sin
+ +
+∫x a a x a a
dxx a
Q-3)(((( )))) (((( ))))
1
cos – .sin –dx
x a x b∫∫∫∫
Ans. Let I = ( ) ( )
1
cos – .sin –dx
x a x b∫
=( )
( )
( ) ( )
1.
cos –
cos –
cos – .sin –
b a
b adx
x a x b∫
= ( )
( ) ( )
( ) ( )
sec – .
cos – – –
cos – .sin –
b a
x a x bdx
x a x b
∫
= ( )
( ) ( )
( ) ( )
( ) ( )
sec –
cos – .cos –
+ sin – .sin –
cos – .sin –
b a
x a x b
x a x bdx
x a x b
∫
= ( )
( )
( )
( )
( )
sec –
cos – sin –+
sin – cos –
b a
x b x adx
x b x a
∫
= ( )
( ) ( )
sec –
cot – + tan –
b a
x b dx x a dx ∫ ∫
= sec(b – a)
[log|sin(x – b) – log|cos(x – a)|] + c
= sec(b – a). log ( )
( )
sin –+
cos –
x bc
x a
Q-4)(((( )))) (((( ))))
1
cos – cos –dx
x a x b∫∫∫∫
Ans. Let I = ( ) ( )
1
cos – cos –dx
x a x b∫
=
( )
( )
( )
cos – .cos2
– sin – .sin2
cos –
x a a
x a adx
x a
∫
= ( )cos2 – tan – sin2a dx x a a dx∫ ∫
= ( )cos2 – sin2 tan –a dx a x a dx∫ ∫= x. cos 2a – sin2 a.log|sec (x – a)|+ c
= ( )
( )
( ) ( )
sin –1
sin – cos – cos –
b adx
b a x a x b∫
= ( )
( ) ( )
( ) ( )
sin – – –1
sin – cos – cos –
x a x bdx
b a x a x b
∫
= ( )
1
sin –b a××××
( ) ( ) ( ) ( )
( ) ( )
sin – cos – – cos – sin –
cos – cos –
x a x b x a x bdx
x a x b∫
= ( )cosec ( – ) tan – – tan( – )b a x a dx x b dx ∫ ∫
= ( )( )
( )
log sec –cosec –
– log sec – +
x ab a
x b c
= ( )( )
( )
sec –cosec – log +
sec –
x ab a c
x b
Indefinite Integration
16 Mahesh Tutorials Science
Q-3)(((( )))) (((( ))))
1
sin – cos –dx
x a x b∫∫∫∫
Ans. Let I = ( ) ( )
1
sin – cos –dx
x a x b∫
∴∴∴∴ I = ( )
( )
( ) ( )
cos –1.
cos – sin – .cos –
b adx
b a x a x b∫
∴∴∴∴ I = ( )
( ) ( )
( ) ( )
cos – – –1
cos – sin – .cos –
x a x bdx
b a x a x b
∫
I = sec(b – a)
( ) ( ) ( ) ( )
( ) ( )
cos – cos – +sin – sin –
sin – .cos –
x a x b x a x bdx
x a x b∫
I = ( ) ( ) ( )sec – cot – + tan –b a x a x b dx ∫I = sec(b – a) [log|sin(x – a)|
+ log|sec(x – b)|] + c
= sec(b – a) [log|sin(x – a)|
– log|cos(x – b)] + c
I = ( )( )
( )
sin –sec – log +
cos –
x ab a c
x b
Q-4) (((( )))) (((( ))))
1
sin – sin –dx
x a x b∫∫∫∫
Ans. = ( ) ( )
1
sin – sin –dx
x a x b∫
= ( )
( )
( ) ( )
sin –1
sin – sin – sin –
a b dx
a b x a x b∫
=( )
( ) ( )
( ) ( )
sin – – –1
sin – sin – sin –
x b x adx
a b x a x b
∫
= ( )
1
sin –a b
( ) ( ) ( ) ( )
( ) ( )
sin – cos – – sin – .cos –
sin – sin –
x b x a x a x bdx
x a x b∫
= ( )
1
sin –a b
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
sin – cos – – sin – .cos –
sin – sin – sin – .sin –
x b x a x a x bdx
x a x b x a x b∫
= ( )( ) ( )
1cot – – cot –
sin –x a x b dx
a b ∫
= ( )
( ) ( )
1
sin –
cot – – cot –
a b
x a dx x b dx ∫ ∫
= ( )
( ) ( )
1
sin
log sin – – log sin – + .
a b
x a x b c
−
= ( )
( )
( )
sin –1log +
sin – sin –
x ac
a b x b
Q-5)(((( ))))cos
cos
++++
∫∫∫∫x a
dxx
Ans. I =( )cos
cos
+
∫x a
dxx
=cos .cos – sin sin
cos∫x a x a
dxx
= [ ]cos – tan sin∫ a x a dx
= cos – tan sin∫ ∫adx x adx
= cos – sin tan∫ ∫a dx a x dx
= cos a (x) – sin a log |sec x| + c
I = x cos a – sin a. log |sec x| + c
Q-6) (((( ))))sin +
sin
x adx
x∫∫∫∫
Ans. Let I =( )sin +
sin
x adx
x∫
=sin .cos + cos .sin
sin
x a x adx
x∫
= ( )cos + cot .sina x a dx∫
= cos + sin cota dx a x dx∫ ∫= x.cos a + sin a. log|sin x| + c
= ( )cos2 – cot sin2+ ∫ a x a a dx
= ( )cos2 – cos sin2+∫ ∫adx x a adx
= ( )cos2 – sin2 cot +∫ ∫a dx a x a dx
I = x cos 2a – sin 2 a log |sin (x + a)| + c
Mahesh Tutorials Science 17
Indefinite Integration
GROUP (C) – CLASS WORK PROBLEMS
Q-1) (((( )))) (((( ))))6
3 –1 +1x x dx∫∫∫∫
Ans. Let I = ( ) ( )6
3 – 1 +1x x dx∫Put x + 1 = t
∴∴∴∴ x = t – 1
∴∴∴∴ dx = dt
∴∴∴∴ I = ( ) 63 –1 –1t t dt ∫
= ( ) 63 – 4t t dt∫
=7 63 – 4t dt t dt∫ ∫
=
8 73 4– +
8 7
t tc
= ( ) ( )8 73 4
+1 – +1 +8 7x x c
Q-2) 3
2 +3
3 – 2
xdx
x∫∫∫∫
Ans. Let I = 3
2 + 3
3 – 2
xdx
x∫Put 3x – 2 = t
∴∴∴∴ dx =1
3dt
∴∴∴∴ I =3
+ 22 + 3
3
3
t
dt
t
∫
= 1
3
1 2 + 4 + 9
33
tdt
t
∫
=
2 –1
3 31
2 +139
t t dt ∫
=
5 –1
3 31 2 13+ +
5 29
3 3
t tc
∫
= ( ) ( )
5 2
3 31 6 393 – 2 + 3 – 2 +
9 5 2x x c
Q-3)5 3 3∫∫∫∫ x a x dx+
Ans. Let I =5 3 3+x a x dx∫
=3 3 3 2+x a x x dx⋅∫
Put a3 + x3 = t
x3 = t –a3
3x2dx = dt
∴∴∴∴ x2dx = 3
dt
∴∴∴∴ I = ( )3–3
dtt a t∫
= 1
3( )3 2 3 1 2–t a t dt∫
= 1
3
5 2 3 3 2
– + c5 3
2 2
t a t
= ( ) ( )
53 2
3 3 3 3 322 + 2 +– +
15 9
a x a a xc
Q-4)(((( ))))
7
241+
∫∫∫∫x
dx
x
Ans. Let I =( )
7
241+
xdx
x∫
=( )
4 3
241+
x xdx
x
⋅
∫
Put 1 + x4 = t ∴∴∴∴ x4 = t – 1
4x3 dx = dt
∴∴∴∴ x3dx = 4
dt
∴∴∴∴ I = 2
1 – 1
4
tdt
t∫
= 2
1 1 1–
4dt
t t
∫
=1 1
log + +4
t ct
∫
=1 1
log + +4 4
t ct
Indefinite Integration
18 Mahesh Tutorials Science
Q-2) (((( ))))
5
22 –x a x dx∫∫∫∫
Ans. Let I = ( )∫ –x a x dx
5
22
Put a – x = t
∴ dx = –dt
∴ I = ( ) ( )∫ – –a t t dt
5
22
= ( )∫– – 2 + .a at t t dt
5
22 2
= ∫– 2 – –at t a t dt
7 9 522 2 2
=
2 – – +9 14 7
2 5 2
t t ta a c
9 17 7
2 5 22
=4 5 2
– – +9 14 7
aat t t c
9 14 72
2 5 2
= ( ) ( )
( )
4 2– – –
9 7
2– – +
11
aa a x a x
a x c
9 72
2 2
11
2
Q-3) 2 + 3
4 – 5
xdx
x∫∫∫∫
Ans. Let I =2 + 3
4 – 5
xdx
x∫Put 4x – 5 = t
∴ dx = 4
dt
GROUP (C) – HOME WORK PROBLEMS
Q-1) (((( )))) (((( ))))5
2 –1 + 4x x dx∫∫∫∫
Ans. Let I = ( ) ( )∫ 2 –1 + 4x x dx5
Put x + 4 = t
∴ dx = dt
∴ I = ( ) ∫ 2 – 4 –1t t dt5
= ( )∫ 2 – 8 –1t t dt5
Q-5)
3
3+
xdx
x x∫∫∫∫
Ans. Let I = ∫ +
xdx
x x
3
3
Put x = t6
∴ dx = 6t5 dt
∴ I = ( )∫ 6+
tt dt
t t
25
3 2
∴ I = ∫6 +1
tdt
t
2
= ( )( )
∫
–16 – + + +1 +
+1t t t t dt
t
4 3 2
= ( )( )
∫ ∫
–16 – + + +1 –
+1t t t t dt dt
t
4 3 2
= ∫6 – + – + – log +1 +
5 4 3 2
t t t tt t c
5 4 3 2
∴ I =
1 1 1 1– + –
5 4 3 26 +
+ – log 1+
x x x x
c
x x
5 2 1 1
6 3 2 3
1 1
6 6
= ( )( )
4
4
1 1log 1+ + +
4 4 1+x c
x
= ( )( )
4
4
1 1log 1+ + +
4 1+x c
x
= ( )∫ 2 – 9t t dt5
= ( )∫ 2 – 9t t dt6 5
=
2 – 9 +7 6
t tc
7 6
= ( ) ( )2 3
+ 4 – + 4 +7 2x x c
7 6
Mahesh Tutorials Science 19
Indefinite Integration
Q-4) 4
– 5
2 –1
xdx
x∫∫∫∫
Ans. Let I = 4
– 5
2 –1
xdx
x∫Put 2x – 1 = t
∴∴∴∴ dx = 2
dt
∴∴∴∴ I =4
+1– 5
2
2
t
dt
t
∫
∴∴∴∴ I =
∫
1 – 9
4
tdt
t
1
4
= ∫
1– 9
4t t dt
3 –1
4 4
=
∫
1– 9 +
7 34
4 4
t tc
7 3
4 4
=
1 4–12 +
4 7t t c
7 3
4 4
= ( ) ( )
1 42 –1 –12 2 –1 +
4 7x x c
7 3
4 4
∴∴∴∴ I = ( ) ( )12 –1 – 3 2 –1 +
7x x c
7 3
4 4
Q-5) (((( ))))
4
2 –1
– 5
xdx
x∫∫∫∫
Ans. Let I =( )∫2 1
– 5
xdx
x4
–
Put x – 5 = t
∴∴∴∴ x = t + 5
∴∴∴∴ dx = dt
∴∴∴∴ I =( ) ∫2 +5 –1t
dtt4
= ∫2 +10 –1t
dtt4
= ∫2 +9t
dtt4
= ∫2 9+ dt
t t3 4
= 2 + 9 +–2 –3
t tc
–2 –3
= –(x – 5)–2 – 3(x – 5)–3 + c
∴∴∴∴ x = +5
4
t
∴∴∴∴ I =
+52 + 3
4
4
t
dt
t
∫
=1 +5 + 6
4 2
tdt
t
∫
=
∫
1+11
8t t dt
1 –1
2 2
=
1+ 11 +
3 18
2 2
t tc
3 1
2 2
= ( ) ( )
1 24 – 5 + 22 4 – 5 +
8 3x x c
3 1
2 2
= ( )
1 114 – 5 + 4 – 5 +
12 4x x c
3
2
Q-6) (((( ))))3 – 2 2 –1x x dx∫∫∫∫
Ans. Let I = ( )3 – 2 2 –1x x dx∫Put 2x – 1 = t
∴∴∴∴ dx = 2
dt
∴∴∴∴ I =( )–1
3 – 22 2
t dtt
∫
= ( )1
3 +3 – 44
t t dt∫
Indefinite Integration
20 Mahesh Tutorials Science
GROUP (D) – CLASS WORK PROBLEMS
Q-1)7 sin + 24cos
3cos + 4 sin
x xdx
x x∫∫∫∫Ans. Let 7 sinx + 24 cos x
= A [3 cosx + 4 sinx] + B [– 3 sinx + 4 cos x]
= (4A – 3B) sin x + (3A + 4B) cos x
∴∴∴∴ 4A – 3B = 7 and 3A + 4B = 24
on solving, we get,
A = 4, B = 3
∴∴∴∴ 7 sin x + 24 cos x = 4 (3 cos x + 4 sinx)
+ 3(4cosx – 3sinx)
∴∴∴∴ I =( )
( )
4 3cos + 4sin+
3cos + 4sin
x xdx
x x∫( )
( )
3 4cos 3sin
3cos 4sin
x xdx
x x
−
+∫=
4cos – 3sin4 + 3
3cos + 4sin
x xdx dx
x x∫ ∫Let f (x) = 4 sinx + 3 cosx
f′′′′ (x) = 4 cosx – 3 sinx
∴∴∴∴ I = 4x + 3log |3 cos x + 4 sin x|+ c
Q-8)
2
3 3
xdx
x a−−−−∫∫∫∫
Ans. I = ∫ –
xdx
x a
2
3 3
Put x3 – a3 = t
(3x2)dx = dt
x2dx = 3
dt
∴∴∴∴ I =1
3∫dt
t
= ∫1
3t dt–1 2
=1
+3 1 2
tc
–1 2
=2
+3t c
= ( )2
– +3
x a c3 3
Q-7) 1 1
2 3
1
+
dx
x x
∫∫∫∫
Ans. Let I = ∫1
+
dx
x x
1 1
2 3
Put x = t6
∴∴∴∴ dx = 6t5 dt
∴∴∴∴ I = ∫6
+
t dt
t t
5
3 2
= ∫6 +1
tdt
t
3
=( )
( )∫–1 –1
6+1
tdt
t
3
= ∫
+1 16 –
+1 +1
tdt
t t
3
=( ) ( ) ∫
+1 – +1 16 –
+1 +1
t t tdt
t t
2
= ∫
16 – +1 –
+1t t dt
t
2
=
6 – + – log +1 +3 2
t tt t c
3 2
=
1 16 – + – log 1+ +
3 2x x x x c1 1 1 1
2 3 6 6
= ∫
13 –
4t t dt
3 1
2 2
=
13 – +
5 34
2 2
t tc
5 3
2 2
= ( ) ( )
1 6 22 – 1 – 2 – 1 +
4 5 3x x c
5 3
2 2
= ( ) ( )3 1
2 –1 – 2 –1 +10 6
x x c
5 3
2 2
Mahesh Tutorials Science 21
Indefinite Integration
Q-2)8sin cos
2sin – 3cos
x x
x x
++++
∫∫∫∫Ans. Let 8 sin x + cos x
= A[2 sinx – 3 cosx] + B[2 cosx + 3 sinx]
= (2A + 3B) sin x + (– 3A + 2B) cos x
∴∴∴∴ 2A + 3B = 8
– 3A + 2B = 1
On solving,we get,
A = 1, B = 2
∴∴∴∴ Nr. = (2 sinx – 3 cosx) + 2 (2 cosx + 3 sinx)
∴∴∴∴ I =2cos + 3sin
1 + 22sin – 3cos
x xdx dx
x x∫ ∫= x + 2 log|2 sin x – 3 cos x| + c
Q-3)1
1+ cotdx
x∫∫∫∫Ans. I =
1
1+ cotdx
x∫=
1
cos1+
sin
dxx
x
∫
=sin
sin + cos
xdx
x x∫=
1 2sin
2 sin + cos
xdx
x x∫
=( ) ( )( )
( )
sin + cos + sin – cos1
2 sin + cos
x x x xdx
x x∫
=( )
( )
cos – sin11. –
2 sin + cos
x xdx dx
x x
∫ ∫
=1
– log sin + cos +2x x x c
Q-4)3 – 4
4 +5
x
x
edx
e∫∫∫∫Ans. Let I = ∫
3 – 4
4 + 5
edx
e
x
x
Let 3ex – 4 = A (4ex + 5) + B d
dx(4ex + 5)
= A(4e2x + 5) + B (4ex )
= ex (4A + 4B) + 5A
∴∴∴∴ 4A + 4B = 3 and
– 5A = –4
∴∴∴∴ A =4
–5
GROUP (D) – HOME WORK PROBLEMS
Q-1) 14 sin +8cos
2sin +3cos
x xdx
x x∫∫∫∫Ans. Let 14 sin x + 8 cos x
= A(2 sin x + 3 cos x)
+ Bd
dx (2 sin x + 3 cos x)
= A(2 sin x + 3 cos x) + B(2 cos x – 3 sin x)
= (2A – 3B) sin x + (3A + 2B) cos x
2A – 3B = 14 ...(i)
3A + 2B = 8 ...(ii)
Solving (i) and (ii) we get
A = 4 and B = –2
14 sin x + 8 cos x
= 4(2 sin x + 3 cos x) –2(2 cos x – 3 sin x)
So,
I =( ) ( )4 2sin + 3cos – 2 2cos – 3sin
2sin + 3cos
x x x xdx
x x∫
∴∴∴∴ I =2cos – 3sin
4 – 22sin + 3cos
x xdx
x x
∫
∴∴∴∴ I = 4x – 2 log|2 sin x + 3 cos x| + c
∴∴∴∴ B =3 4 31
+ =4 5 20
∴∴∴∴ I =
( ) ( )
( )∫4 31
– 4 + 5 + 4 + 55 20
4 + 5
de e
dx
e
x x
x
=( )( )
( )
( )∫ ∫4 + 54 + 54 31
– + .5 204 + 5 4 + 5
dee
dxdx dxe e
xx
x x
= ∫4 31
– + log 4 + 5 +5 20
dx e cx
=4 31
– + log 4 + 5 +5 20x e cx
Indefinite Integration
22 Mahesh Tutorials Science
Q-4)20 –12
3 – 4
x
x
edx
e∫∫∫∫
Ans. = ∫10 – 6
23 – 4
edx
e
x
x
=( )∫
6 –10–2
3 – 4
edx
e
x
x
= ∫6 – 8 2
–2 –3 – 4 3 – 4
edx
e e
x
x x
= ∫ ∫
2–2 2 –
3 – 4
edx dx
e
–
–
x
x
Let 3 – 4e–x = f (x)
∴∴∴∴ f′ (x) = 4e–x
= ∫ ∫
1 4–2 2 –
2 3 – 4
edx dx
e
–
–
x
x
=
1–2 2 – log 3 – 4 +
2x e c–x
= log |3 – 4e–x| – 4x + c
= log |3ex – 4| – log ex – 4x + c
= log |3ex – 4| – x log e – 4x + c
= log |3ex – 4| – 5x + c ..... (∵∵∵∵ log e = 1)
Q-3)3 + 4
2 – 8
x
x
edx
e∫∫∫∫
Ans. Let I = ∫3 + 4
2 – 8
edx
e
x
x
Let 3ex + 4 = A(2ex – 8) + Bd
dx(2ex – 8)
∴∴∴∴ 3ex + 4 = A(2ex – 8) + B(2ex)
∴∴∴∴ 3ex + 4 = (2A + 2B) ex – 8A
∴∴∴∴ 2A + 2B = 3 ...(i)
–8A = 4 ...(ii)
Solving (i) and (ii) we get
A = 1
–2
and B = 2
∴∴∴∴ 3ex + 4 = 1
–2
(2x – 8) + 2(2ex)
I =( ) ( )
∫1
– 2 – 8 + 2 22
2 – 8
e e
dxe
x x
x
I =( )
∫
21– + 2
2 2 – 8
edx
e
x
x
I =1
–2x + 2log|2ex – 8| + c
Q-2)1
1 tandx
x∫∫∫∫ ++++
Ans. I =1
1+ tandx
x∫=
1
sin1+
cos
dxx
x
∫
=cos
cos + sin
xdx
x x∫=
1 2cos
2 cos + sin
xdx
x x∫
=( ) ( )( )
( )
cos + sin + cos sin1
2 cos + sin
x x x xdx
x x
−
∫
=( )
( )
cos – sin11. +
2 sin + cos
x xdx dx
x x
∫ ∫
=1
+ log sin + cos +2x x x c
Mahesh Tutorials Science 23
Indefinite Integration
GROUP (E) – CLASS WORK PROBLEMS
Q-1) 2
1
4 + 9dx
x∫∫∫∫Ans. Let I = ∫
1
4 + 9dx
x2
= ( ) ( )∫1
2 + 3dx
x22
=
1 2 1tan +
3 3 2
xc–1 ××××
=
1 2tan +
6 3
xc–1
Q-2) 2
sec .tan
9 –16 sec
x xdx
x∫∫∫∫
Ans. Let I =sec .tan
9 –16sec
x xdx
x∫ 2
Put sin x = t
∴∴∴∴ sec x . tan x dx = dt
∴∴∴∴ I = ∫ 9 –16dt
t2
= ( ) ( )∫ 3 – 4
dt
t2 2
= ××××××××
1 3+ 4 1log +
2 3 3 – 4 4
tc
t
=1 3+4seclog +
24 3 – 4sec
xc
x
Q-5) 2
1
+6 +10dx
x x∫∫∫∫Ans. Let I = ∫
1
+6 +10dx
x x2
= ∫1
+6 +9+1dx
x x2
Q-3) 2
cos
10 – cos∫∫∫∫x
dx
x
Ans. Let I = 2
cos
10 – cos
xdx
x∫
= ( )2
cos
10 – 1 – sin
xdx
x∫
I = 2
cos
9 + sin
xdx
x∫Put sin x = t
∴∴∴∴ cosx dx = dt
∴∴∴∴ I =( )
2 2
1
3 + tdt∫
=( )
22
1
t + 3dt∫
I =2log t + t + 9 + c
I =2log sin + sin + 9 +x x c
=2log sin + 1 – cos + 9 +x x c
I =2log sin + 10 – cos +x x c
Q-4) –
1
4 + 9e∫∫∫∫ x xdx
e
Ans. Let I = –x x
1
4 + 9edx
e∫
= 2
x
x4 +9
edx
e∫Put ex = t ∴∴∴∴ exdx = dt
I = 2
1
4 + 9dt
t∫
=( )
2 2
1
2 +3dt
t∫
= ( )22
1 1
4 + 3 2dt
t∫
=–11 1
tan +34 3 2
2
tc
=–11 2
tan +6 3
tc
I =–1
x1 2tan +
6 3
ec
Indefinite Integration
24 Mahesh Tutorials Science
Q-9)
2
2
sec
5tan –12 tan +14
xdx
x x∫∫∫∫
Ans. Let I = ∫sec
5 tan –12tan +14
xdx
x x
2
2
Put tan x = t
sec2 x dx = dt
∴∴∴∴ I = ∫ 5 –12 +14
dt
t t2
= ∫1
12 145– +5 5
dt
t t2
= ∫1
12 144 14 1445– + + –5 100 5 100
dt
t t2
=
∫1
5 12 2 34– +10 10
dt
t
22
=
12–
1 1 10tan +5 2 342 34
1010
t
c–1
××××
=
1 10 –12tan +
34 2 34
tc
–1
=
1 5 tan – 6tan +
34 34
xc
–1
Q-6) 2
1
15+4 – 4dx
x x∫∫∫∫
Ans. Let I = ∫1
15 + 4 – 4dx
x x2
=( )∫15 – 4 – 4 +1 +1
dtdx
x x2
=( ) ( )∫ 4 – 2 –1
dtdx
x2 2
= ××××××××
1 4+ 2 –1 1log +
2 4 4 – 2 +1 2
xc
x
=1 3+2log +
16 5 – 2
xc
x
Q-7) 2
1
+8 +25dx
x x∫∫∫∫
Ans. Let I = ∫1
+8 + 25dx
x x2
= ∫1
+ 8 +16 + 9dx
x x2
=( ) ( )∫
1
+ 4 + 3dx
x2 2
= ( ) ( ) ( )log + 4 + + 4 + 3 +x x c2 2
= ( )log + 4 + + 8 + 25 +x x x c2
Q-8) 2
1
3 – 4 +2dx
x x∫∫∫∫
Ans. Let I = ∫1
3 – 4 + 2dx
x x2
=1 1
3 4 2– +3 3
dx
x x∫
2
=1 1
3 4 4 2 4– + + –3 9 3 9
dx
x x∫
2
= ( )∫1
+ 3 +1dx
x2 2
= tan–1 (x + 3) + c
=1 1
32 2
– +3 3
dx
x
∫ 22
=
1log
3
2 2 2– + – – +3 3 3
x x c
22
=1 2 4 2log – + – + +
3 33x x x c
x
2
Mahesh Tutorials Science 25
Indefinite Integration
Q-10) 2 +4 +13
x
x x
edx
e e∫∫∫∫
Ans. Let I = ∫ +4 +13
edx
e e2
x
x x
put ex = t
∴∴∴∴ ex dx = dt
∴∴∴∴ I = ∫ + 4 +13
dt
t t2
= ∫ + 4 + 4 + 9
dt
t t2
=( ) ( )∫ +2 2+ 3
dt
t2
= ( ) ( ) ( )log + 2 + + 2 + 3 +t t c2 2
= ( ) ( ) ( )log + 4 + + 2 + 3 +e e c2x t
= ( )log + 2 + + 4 +13 +e e e cx 2x x
Q-11)2sin cos – 2cos 2x x x dx++++∫∫∫∫
Ans. I = ∫sin cos – 2cos +2x x x dx2
Put cos x = t ∴∴∴∴ –sin x dx = dt
I = ∫– – 2 +2t t dt2
= ( )∫– –1 +1t dt2 2
=
( )
( )
–1–1 +1
2– +
1+ log –1+ –1 +122
tt
c
t t
2 2
2
=
cos –1– cos – 2cos2+2
2
xx2
1+ log cos –1+ cos – 2cos2+ 2 +2
x x c2
=1 – cos
cos – 2cos + 22
xx x2
1
– log cos –1+ cos – 2cos + 2 +2
x x x c2
Q-12) I =2/3 2/3
1
– 4dx
x x∫∫∫∫Ans. I = Put x1/3 = t
∴∴∴∴–2/31
3x dx dt=
∴∴∴∴ –2/3 3x dx dt=
I =2 2
13
– 2dt
t∫=
2 23 log + – 2 +t t c
I =1/3 2/33 log + – 4 +x x c
Q-13) I =sin cos
7 – 9sin2
++++
∫∫∫∫x x
dxx
Ans. Let I = –9
sec + cos
7 sin2
x xdx
x∫Put t = sin x – cos x
∴∴∴∴ dt = (cos x + sin x)dx
and t2 = sin2x + cos2x – 2 sin x.cos x
∴∴∴∴ t2 = 1 – sin 2x
∴∴∴∴ sin 2x = 1 – t2
∴∴∴∴ I = ( )27 – 9 1 –
dt
t∫
= 27 – 9+ 9
dt
t∫= 29 – 2
dt
t∫
= ( ) ( )22
3 – 2
dt
t∫
= ××××1 3 – 2 1
log +32 2 3 + 2
tc
t
∴∴∴∴ I =( )
( )
3 sin – cos – 21log +
6 2 3 sin – cos + 2
x xc
x x
Indefinite Integration
26 Mahesh Tutorials Science
GROUP (E) – HOME WORK PROBLEMS
Q-1) 2
1
+6 +25dx
x x∫∫∫∫
Ans. Let I = 2
1
+6 +25dx
x x∫= 2
1
+6 +9+16dx
x x∫
=( ) ( )
2 2
1
+3 + 4dx
x∫
=–11 +3
tan +4 4
xc
Q-2) 2
1
9 +6 +5dx
x x∫∫∫∫
Ans. Let I = 2
1
9 +6 +5dx
x x∫=
2
1 1
2 59– +3 9
dx
x x∫
Q-14) cos + sin
7 – 2cos2
x xdx
x∫∫∫∫Ans. Let I = –2
cos +sin
7 cos2
x xdx
x∫=
cos sin+
7 – 2cos2 7 – 2cos2
x xdx dx
x x∫ ∫
=( )
( )
2
2
cos
7 – 2 1 – 2sin
sin+
7 – 2 2cos –1
xdx
x
xdx
x
∫
∫
= 2 2
cos sin+
5 +4sin 9 – 4cos
x xdx dx
x x∫ ∫Put sin x = t
1, in first integral
∴∴∴∴ cos x dx = dt1 and
cos x dx = t2 in second integral
∴∴∴∴ sin x dx = –dt2
∴∴∴∴ I =21
2 21 2
–+
5 + 4 9 – 4
dtdt
t t∫ ∫
=( ) ( ) ( ) ( )
21
2 22221
–3 – 25 + 2
dtdt
tt∫ ∫
=–1 1
2
2
××××
××××××××
21 1tan
25 5
3+ 21 1– log +2 3 3 – 2 2
t
tc
t
∴∴∴∴ I =–11 2sin
tan2 5 5
1 3 +2cos– log +16 3 – 2cos
x
xc
x
Q-15)cos – sin
8 – sin2∫∫∫∫x x
dxx
Ans. I =( )
cos – sin
9 – 1 sin2
x xdx
x+∫
= ( )2 2
cos – sin
9 – cos + sin + 2sin cos
x xdx
x x x x∫
=( )
22
cos – sin
3 – cos + sin
x xdx
x x∫Put sin x + cos x = t
∴∴∴∴ (cos x – sin x) dx = dt
Q-16) cos cos2∫∫∫∫ x x dx
Ans. I =2cos 1 – 2sinx x dx∫
Put sin x = t cos x dx = dt
I =21– 2t dt∫
=21
4 –2
t dt∫
=
2
214 –
2t dt
∫
= 2 –11 1/ 2
4 – + sin +2 2 2 1/ 2
t tt c
I = ( )–1sin 1cos2 + sin 2sin +
2 2 2
xx x c
∴∴∴∴ I =–1
2 2
1= sin +
33 –
tdt c
t
∫
∴∴∴∴ I =–1 sin + cos
sin +3
x xc
Mahesh Tutorials Science 27
Indefinite Integration
Q-3) 2
1
9+8 +dx
x x∫∫∫∫
Ans. Let I = 2
1
9+8 +dx
x x∫
= ( )2
1
9 – – 8 +16 +16x x∫
= ( ) ( )2 2
1
5 – – 4dx
x∫
=–1 – 4
sin +5
xc
Q-4) 2
1
2 – 4 +7dx
x x∫∫∫∫Ans. Let I =
2
1
2 – 4 + 7dx
x x∫=
1 1
2 7– 2 +
2
dx
x x∫
2
=1 1
2 7– 2 +1+ –1
2
dx
x x∫ 2
=
( )
1 1
25
–1 +2
dx
x
∫ 2
2
= ( ) ( )
2
21 5log –1 + –1 + +
22x x c
= ( ) 21 7log –1 + – 2 + +
22x x x c
Q-5)
(((( ))))2
1
log + 4dx
x x∫∫∫∫
Ans. Let I =
( )2
1
log + 4dx
x x∫
Put log x = t ∴∴∴∴1
xdx = dt
∴∴∴∴ I =( ) ( )
2 2 2=
+ 4 + 2
dt dt
t t∫ ∫
= ( )22log + + 2 +t t c
= ( )2
log log + log + 4 +x x c
Q-6) 2
cos
sin – 2sin +5
xdx
x x∫∫∫∫
Ans. Let I = 2
cos
sin – 2sin +5
xdx
x x∫Put sin x = t
∴∴∴∴ cos x dx = dt
∴∴∴∴ I =2 – 2 +5
dt
t t∫
=2 – 2 +1+ 4
dt
t t∫
=( ) ( )
2 2–1 + 2
dt
t∫
= ( ) ( ) ( )2 2
log –1 + –1 + 2 +t t c
= ( ) 2log –1 + – 2t + 5 +t t c
= ( ) 2log sin –1 + sin – 2sin +5 +x x x c
=2
1 1
2 1 5 19– + + –3 9 9 9
dx
xx∫
= 2 2
1 1
9 1 2+ +3 3
dx
x
∫
= –1
11 1 3tan +2 29
3 3
x
c
+
=–11 3 +1
tan +6 2
xc
Q-7) 21–
x
x
adx
a∫∫∫∫Ans. put ax = t
ax log a dx = dt
ax dx = log
dt
a
Indefinite Integration
28 Mahesh Tutorials Science
Q-10) (((( )))) (((( ))))2
1
1+ cos2 1 – tandx
x x∫∫∫∫
Ans. Let I =( ) ( )2
1
1+ cos2 1 – tandx
x x∫
= ( ) ( )
2
2 2 2
cos
2cos cos – sin
xdx
x x x∫
=1
sec22
x dx∫=
1log sec2 + tan2 +
4x x c
Q-11) sin + cos
3 +4 sin2
x xdx
x∫∫∫∫
Ans. Let I =sin + cos
3+ 4sin2
x xdx
x∫Put sin x – cos x = t
∴∴∴∴ (cos x + sin x)dx = dt
∴∴∴∴ t2 = sin2x + cos2x – 2 sin x.cos x
t2 = 1 – sin 2x
∴∴∴∴ sin 2x = 1 – t2
∴∴∴∴ I = ( )23+ 4 1 –
dtdx
t∫
= 27 +4
dt
t∫=
( ) ( )2 2
7 – 2
dt
t∫
=( )
××××1 7 + 2 1
log +27 – 22 7
tc
t
∴∴∴∴ I =( )
( )
7 +2 sin – cos1log +
4 7 7 – 2 sin – cos
x xc
x x
Q-12) cos + sin
3 – 2cos2
x xdx
x∫∫∫∫
Ans. Let I =cos +sin
3 – 2cos2
x xdx
x∫=
cos sin+
3 – 2cos2 3 – 2cos2
x xdx dx
x x∫ ∫
Q-8)
2
2
sec
9 – 5tan
xdx
x∫∫∫∫
Ans. Let I =
2
2
sec
9 – 5 tan
xdx
x∫Put tan x = t
∴∴∴∴ sec2 x dx = dt
∴∴∴∴ I = 29 – 5
dt
t∫
=( ) ( )
2
3 – 5
dt
t∫
= ××××××××
1 3+ 5 1log +
1 3 3 – 5 5
tc
t
=1 3+ 5 tan
log +6 5 3 – 5 tan
xc
x
Q-9) (((( ))))
2
1
log – 9dx
x x
∫∫∫∫
Ans. Let I =( )
2
1
log – 9dx
x x
∫Put log x = t
∴∴∴∴1
xdx = dt
∴∴∴∴ I = 2 – 9
dt
t∫
= ( ) ( )2 2– 3
dt
t∫
=××××
1 – 3log +
1 3 +3
tc
t
=1 log – 3log +
6 log + 3
xc
x
=2
log
1 –
dt
a
t∫
= ( )–11sin +
logt c
x
= ( )–1 x1sin +
loga c
a
Mahesh Tutorials Science 29
Indefinite Integration
Q-13)24 5x dx−−−−∫∫∫∫
Ans. I =24 5x dx−∫
=2 5
2 –4
x dx∫
=
2
2 52 –
4x dx
∫
=
2
2
2–
2
2
5
25–
2 2 22
5log
2
xx
c
x x
+ +
+ +
=2 25
4 – 5 – log 2 4 – 52 4
xx x x c+ +
Q-14)2 4 5x x dx+ ++ ++ ++ +∫∫∫∫
Ans. I =2 +4 +5x x dx∫
=2 +4 +4+1x x dx∫
= ( )2
+2 +1x dx∫
= ( )2
2
+ 2+ 2 +1
2
1+ log + 2+ + 4 +5 +2
xx
x x x c
I =2
2
+ 2+ 4 +5
2
1+ log + 2+ + 4 +5 +2
xx x
x x x c
Q-15)29 6 7x x dx+ ++ ++ ++ +∫∫∫∫
Ans. I =29 +6 +7x x dx∫
=2 6 7
3 + +9 9
x x dx∫
=2 2 1 2
3 + + +3 9 3
x x dx∫
=
221 2
3 + +3 3
x dx ∫
=
( )
22
2
1+
1 223 + +
2 3 3
2 1 2 7+ log + + + + +3 3 3 3 9
x
x
xx x c
I =2
2
3 +1 2 7+ +
2 3 9
1 2 7+ log + + + + +
3 3 9
x xx
xx x c
=( )
( )
2
2
cos
3 – 2 1 – 2sin
sin+
3 – 2 2cos –1
xdx
x
xdx
x
∫
∫
= 2 2
cos sin+
1+ 4sin 5 – 4cos
x xdx dx
x x∫ ∫Put sin x = t
1, in first integral
∴∴∴∴ cos x dx = dt1 and
Put cos x = t2 in second integral
∴∴∴∴ sin x dx = –dt2
∴∴∴∴ I =21
2 21 2
–5 + 4 5 – 4
dtdt
t t∫ ∫
= ( ) ( ) ( ) ( )
21
2 2 2 21 2
–1 + 2 5 – 2
dtdt
t t∫ ∫
=( )–1
1
2
2
××××
tan 2
2
5 + 21 1– log +
22 5 5 – 2
t
tc
t
= ( )–11tan 2sin
2
1 5 + 2cos– log +4 5 5 – 2cos
x
xc
x
Indefinite Integration
30 Mahesh Tutorials Science
GROUP (F) – CLASS WORK PROBLEMS
Q-1) 2
5 –1
2 +3 + 7
xdx
x x∫∫∫∫
Ans. Let I = 2
–5 1
2 +3 +7
xdx
x x∫d
dx (2x2 + 3x + 7) = 4x + 3
Let 5x – 1 = A(4x + 3) + B. Then 4A = 5
and 3A + B = – 1
∴∴∴∴ A = 5
2and
53
4
+ B = –1
∴∴∴∴ B = 19
–4
∴∴∴∴ I =( )
2
5 194 + 3 –
4 4
2 + 3 + 7
x
dxx x
∫
= 2 2
5 4 +3 19–
4 42 +3 +7 2 +3 +7
x dxdx
x x x x∫ ∫= 2
2
5log 2 +3 +7
4
19 1–
3 74 2+ +2 2
x x
dx
x x
×
∫
... (i)
Now,
=2 3 7+ +2 2
dx
x x∫
=2 3 9 7 9+ + + –2 16 2 16
dx
x x
∫
=22
3 47+ +4 4
dx
x x
∫
= –1
3+
1 4tan
47 47
4 4
x
=–14 4 + 3
tan47 47
x
.... (ii)
from (i) and (ii)
I = 2 –15 19 4 +3
log 2 + 3 + 7 – tan4 47 47
xx x
(from i and ii)
Q-2) 2
2 5
+ 4 +5
xdx
x x
++++
∫∫∫∫
Ans. Let I = 2
2 +5
+ 4 +5
xdx
x x∫
=2 + 4 1
++ 4 +5 + 4 +5
xdx
x x x x
∫ 2 2
= 2 2
2 + 4 1+
+ 4 +5 + 4 +5
xdx
x x x x∫ ∫
Q-16)25 – 4 – 3x x x
e e e dx∫∫∫∫
Ans. I =2x x x5 – 4 – 3e e e dx∫
Put ex = t
ex dx = dt
I =25 – 4 – 3t t dt∫
=2 4 3
5 – –5 5
tt dt ∫
=2 2 4 19
5 – 2 + –5 25 25
t t dt ∫
=
( )
2
22
22
19
5– 2 5 2 19– – –
2 5 5 25 +
2 2 19log – + – –
5 5 5
tt
c
t t
=
2
2
x x
x
x
x x
5 – 2 4 3– –
10 5 55 +
19 2 4 3– log – – –
5 5 550
e ee
ce
e e
+
Mahesh Tutorials Science 31
Indefinite Integration
Q-3) (((( )))) 23 – 2 + +1x x x dx∫∫∫∫Ans. Let I = ( )3 – 2 + +1x x x dx∫ 2 ... (i)
3x – 2 = ( )2. + +1 +d
A x x Bdx
= A(2x + 1) + B ... (ii)
∴∴∴∴ 3x – 2 = (2A)x + (A + B)
comparing coefficients of x and constant term
on both sides,
2A = 3 and A + B = – 2
∴∴∴∴ A = 3
2 and B =
7–2
...(iii)
∴∴∴∴ From (i), (ii) and (iii)
I = ( )3 72 +1 – + +1
2 2x x x dx
∫ 2
= ( )3
2 +1 + +12
7– + +12
x x x dx
x x dx
∫∫
2
2
.....(iv)
Let I1= ( )2 +1 + +1x x x dx∫ 2
Put x2 + x + 1 = t
∴∴∴∴ (2x + 1) dx = dt
I1= .∫ t dx
=
1
2t dt∫
=
3
2+
3
2
tc
∴∴∴∴ I1= ( )
32 2
1
2+ +1 +
3x x c
Let I2=
2 + +1x x dx∫
=2 1 3+ + +
4 4x x dx∫
=
221 3
+ +2 2
x ∫
= 2
2
2
1 1+ + +1
2 2
3
2 1+ log + + + +1
2 2
x x x
x x x
∴∴∴∴ I1= ( ) ( )
32 22
2
2 7+ +1 – 2 +1 + +1
3 8
21 1– log + + + +1 +16 2
x x x x x
x x x c
=
( ) ( )
2
2 2
12 + 4 +5 +
+2 + 2x x dx
x∫
∵∵∵∵ =( )
( )
f xdx
f x
′
∫= ( )2 +f x c
=
( ) ( ) ( )
2
2 2
2 + 4 +5
+ log + 2 + + 2 + 2 +
x x
x x c
=
( )
2
2
2 + 4 +5
+ log + 2 + + 4 +5 +
x x
x x x c
Q-4)+1
9 –
xdx
x∫∫∫∫Ans. I =
+1
9 –
xdxx∫
= ××××+1 +1
9 – –1
x xdx
x x∫ ∫
= ( ) ( )
+1
9 – +1
xdx
x x∫
= 2
+1
8 – + 9
xdx
x x∫Let x + 1 = A
d
dx (8x – x2 + 9)+ B
= A (8 – 2x) + B ..... (ii)
= (8A + B) – 2Ax.
∴∴∴∴ 8A + B = 1, – 2A = 1
Indefinite Integration
32 Mahesh Tutorials Science
Q-3) 2
+3
+2 +2
xdx
x x∫∫∫∫
Ans. Let I = 2
+ 3
+ 2 + 2
xdx
x x∫ ...(i)
Let x + 3
= A d
dx (x2 + 2x + 2)+ B
= A (2x + 2) + B ...(ii)
= 2Ax + (2A + B)
GROUP (F) – HOME WORK PROBLEMS
Q-1) 2
+ 4
+5
xdx
x∫∫∫∫
Ans. Let I = 2
+ 4
+5
xdx
x∫= 2 2
1 1+4
+5 +5dx dx
x x∫ ∫=
( ) ( )2 2
2
1 2 1+ 4
2 +5 + 5
xdx
x x∫ ∫
=2 –11 4
log +5 + tan +2 5 5
xx c
Q-2) 2
5 –1
3 + +2
xdx
x x∫∫∫∫Ans. Let I =
2
5 –1
3 + +2
xdx
x x∫ ...(i)
Let 5x + 1
= A d
dx (3x2 + x + 2)+ B
= A (6x – 1) + B ...(ii)
= 6Ax (A + B)
∴∴∴∴ 6A = 5
∴∴∴∴ A = 5
6 and
∴∴∴∴ A + B = –1
∴∴∴∴ B = –1 – A
∴∴∴∴ B = –1 – 5
6 =
11–6
...(iii)
∴∴∴∴ From (i) and (iii),
I =( )
2
5 116 +1 –
6 6
3 + + 2
xdx
x x∫= 2 2
5 6 +1 11–
6 63 + +2 3 + +2
x dxdx
x x x x∫ ∫= 2
2××××
5 11log 3 + + 2 –
26 6 3+ +3 3
dxx x
xx∫
= 2
2
5log 3 + + 2
6
11–
1 1 218+ + – +3 36 36 3
x x
dx
xx∫
= 2
22
5log 3 + +2
6
11–18 1 23
+ +6 6
x x
dx
x
∫
=2
–1
5log 3 + + 2
6
11 6 +1– tan +3 23 23
x x
xc
∴∴∴∴ A = –1
2
∴∴∴∴ B = 1 – 8A = 1–(–4) = 5
∴∴∴∴ x + 1 –1
2 (8 – 2x) + 5 ...(ii)
From (i) and (ii),
I =( )
2
18 – 2 +5
2
8 – + 9
x
dxx x∫
=2
2
1 8 – 2–2 8 – + 9
+58 – + 9
xdx
x x
dx
x x
∫∫
= ( )
( )
2
2
1– 2 8 – + 92
+525 – – 8 +16
x x
dx
x x
∫∫
=( ) ( )
2
2 2– 8 – + 9 +5
5 – – 4
dxx x
x∫
=2 –1 – 4
– 8 – +9 +5sin +5
xx x c
Mahesh Tutorials Science 33
Indefinite Integration
Q-4)+1
+3
xdx
x∫∫∫∫Ans. Let I =
+1
+ 3
xdx
x∫ =+1 +1
.+ 3 +1
x xdx
x x∫=
( ) ( )
+1
+3 +1
xdx
x x∫
I = 2
+1
+ 4 + 3
xdx
x x∫ ...(i)
Let
x + 1 = A d
dx (x2 + 4x + 3)+ B
= A (2x + 4) + B ...(ii)
= 2Ax + (4A + B)
∴∴∴∴ 2A = 1 ⇒⇒⇒⇒ A = 1
2
∴∴∴∴ 4A + B = 1 ⇒⇒⇒⇒ B = 1 – 4A = 1 – 2 = –1
∴∴∴∴ From (i) and (ii)
∴∴∴∴ I =( )
2
12 + 4 –1
2
+ 4 +3
x
dxx x∫
∴∴∴∴ I = 2 2
1 2 + 4–
2 + 4 + 3 + 4 + 3
x dxdx
x x x x∫ ∫
= ( )( ) ( )
2
2 2
12 + 4 +3 +
2 +1 – 1
dxx x
x∫
=
( ) ( )
2
2 2
+ 4 +3
+ log + 2+ +1 + 1 +
x x
x x c
= 2
2
+ 4 + 3
– log + 2+ + 4 + 3 +
x x
x x x c
Q-5)+ 4
3 –
xdxx∫∫∫∫
Ans. Let I =+ 4
3 –
xdxx∫ = ××××
+ 4 + 4
3 – + 4
x xdx
x x∫=
( ) ( )
+ 4
3 – + 4
xdx
x x∫
= 2
+ 4
– +12
xdx
x x∫ ...(i)
Let
x + 4 = A d
dx (x + x2 + 12)+ B
= A (1 – 2x) + B ...(ii)
= (A + B) – 2Ax
∴∴∴∴ A + B = 4 and – 2A = 1
∴∴∴∴ A = 1
–2
∴∴∴∴ B = 4 – A = 4–1
–2
= 9
2
∴∴∴∴ From (i) and (ii)
∴∴∴∴ I =( )
2
1 9– 1 – 2 +2 2
+ +12
x
dxx x∫
=1 1 – 2
–2 – +12
xdx
x x∫ 2
9+2 – +12
dxdx
x x∫ 2
= ( )2
2 2
1 92 – +12 +
2 2 7 1– –
2 2
dxx x
x
∫
=2 –1
1–
9 2– +12 + sin +72
2
x
x x c
=2 –19 2 –1
– +12 + sin +2 7
xx x c
∴∴∴∴ 2A = 1 ⇒⇒⇒⇒ A = 1
2
∴∴∴∴ 2A + B = 3 ⇒⇒⇒⇒ B = 3 – 2A = 3 – 1
∴∴∴∴ B = 2 ...(iii)
∴∴∴∴ From i), ii) and iii)
∴∴∴∴ I =( )
2
12 +2 +2
2
+2 +2
x
dxx x∫
=2 2
1 2 + 2+ 2
2 +2 + 2 + 2 + 2
x dxdx
x x x x∫ ∫= ( )
( ) ( )
2
2 2
12 +2 + 2 +2
2 +1 + 1
dxx x
x∫
=
( ) ( ) ( )
2
2 2
+2 +2
+2log +1 + +1 + 1 +
x x
x x c
=
( )
2
2
+ 2 + 2
+ 2log +1 + + 2 + 2 +
x x
x x x c
Indefinite Integration
34 Mahesh Tutorials Science
GROUP (G) – CLASS WORK PROBLEMS
Q-1) 2
1
4 +3sindx
x∫∫∫∫
Ans. Let I = 2
1
4+ 3sindx
x∫
=( ) ( )
22
1
2 3 sin
dx
x∫
=–11 3sin 1
tan × +2 2 3
xc
=–11 3 sin
tan +22 3
xc
Q-2)cos
cos3
xdx
x∫∫∫∫
Ans. Let I =cos
cos3
xdx
x∫
= 3
cos
4cos – 3cos
x dx
x x∫= 24cos – 3
dx
x∫Divding both num. and deno. by cos2 x,
we get,
I =
2
2
sec
4 – 3sec
x dx
x∫
= ( )
2
2
sec
4 – 3 1+ tan
x dx
x∫
=
2
2
sec
1 – 3 tan
x dx
x∫Put tan x = t. Then sec2 x dx = dt
∴∴∴∴ I = 21 – 3
dt
t∫=
2
2
1
3 1–
3
dt
t
∫
=
1+
1 1 3. log +
3 1 12 –
3 3
t
c
t
=1 1+ 3 .tan
log +2 3 1 – 3 .tan
xc
x
Q-3) 2 2
1
2sin – 3cos + 7dx
x x∫∫∫∫Ans. I = 2 2
1
2sin – 3cos +7dx
x x∫Dividing Nr and Dr by cos2x
∴∴∴∴ I =
2
2 2
sec
2tan – 3+ 7sec
xdx
x x∫
= ( )
2
2 2
sec
2tan – 3 + 7 1+ tan
xdx
x x∫
=
2
2
sec
9tan + 4
x dx
x∫put tan x = t
∴∴∴∴ sec2 x dx = dt
∴∴∴∴ I = 29 + 4
dt
t∫ =
( ) ( )2 2
3 + 2
dt
t∫
=–1
××××1 3 1tan +
2 2 3
tc
=–11 3tan
tan +6 2
xc
Q-4) 2 2
1
sin + tandx
x x∫∫∫∫
Ans. I = 2 2sin + tan
dx
x x∫
=
2
2 2 2
sec
tan + tan sec
x dx
x x x∫
= ( )
2
2 2 2
sec
tan + tan 1+ tan
x dx
x x x∫
=
2
4 2
sec
tan + 2tan
x dx
x x∫Put tan x = t
Mahesh Tutorials Science 35
Indefinite Integration
∴∴∴∴ sec2 x dx = dt
I = 4 2+2
dt
t t∫
= ( )2 2 +2
dt
t t∫
= ( )
2 2
2 2
1 +2 –
2 + 2
t tdt
t t∫
= 2 2
1 1–
2 + 2
tdt dt
t t
∫ ∫
= 2 22
1 1 1–
2 2dt dt
t t
∫ ∫ +
=–11 1 1
– – tan +2 2 2
tc
t
=–11 1 1 tan
– – tan +2 tan 2 2
xc
x
GROUP (G) – HOME WORK PROBLEMS
Q-1) 2
1
4 +5sindx
x∫∫∫∫
Ans. Let I = 2
1
4+5sindx
x∫Dividing Nr and Dr by cos2x
∴∴∴∴ I =
2
2 2
sec
4sec + 5 tan
xdx
x x∫
= ( )
2
2 2
sec
4 tan +1 +5 tan
xdx
x x∫
=
2
2
sec
4 + 9tan
xdx
x∫put tan x = t
∴∴∴∴ sec2 x dx = dt
∴∴∴∴ I = 24+9
dt
t∫ =
( ) ( )2 2
2 + 3
dt
t∫ = –1
××××1 3 1tan +
2 2 3
tc
=–11 3tan
tan +6 2
xc
Q-3) 2 2
1
sin +2cos +3dx
x x∫∫∫∫
Ans. Let I = 2 2
1
sin +2cos + 3dx
x x∫Dividing Nr and Dr by cos2x
∴∴∴∴ I =
2
2 2
sec
tan + 2+ 3sec
xdx
x x∫
= ( )
2
2 2
sec
tan + 2+ 3 1+ tan
xdx
x x∫
=
2
2
sec
5 + 4tan
xdx
x∫put tan x = t
∴∴∴∴ sec2 x dx = dt
Q-2) 2
1
9 16cosdx
x++++∫∫∫∫
Ans. I = 2
1
9 16cosdx
x+∫Dividing numerator and denominator
by cos2x
I =
2
2
sec
9sec +16
xdx
x∫
= ( )
2
2
sec
9 1+ tan +16
xdx
x∫
I =
2
2
sec
9tan +25
xdx
x∫Put tan x = t
∴∴∴∴ sec2 x dx = dt
I = 2
1
9 + 25dt
t∫
= 2
2
1 1
9 5+
3
dt
t
∫
= ( )–11 1
tan +9 5 3 5 3
tc
=–11 3
tan +15 5
tc
=–11 3tan
tan +15 5
xc
Indefinite Integration
36 Mahesh Tutorials Science
Q-4) 2 2
1
4cos + sindx
x x∫∫∫∫Ans. Let I = 2 2
1
4cos + sindx
x x∫Dividing Nr and Dr by cos x
∴∴∴∴ I =
2
2
sec
4 + tan
xdx
x∫put tan x = t
∴∴∴∴ sec2 x dx = dt
∴∴∴∴ I = 24 +
dt
t∫ =( ) ( )
2 22 +
dt
t∫
=–11
tan +2 2
tc
=–11 tan
tan +2 2
xc
∴∴∴∴ I = 25 + 4
dt
t∫
= ( ) ( )2 2
5 + 2
dt
t∫
=–1
××××1 2 1
tan +25 5
tc
=–11 2tan
tan +2 5 5
xc
Q-5) 2 2
1
9cos 16sindx
x x++++∫∫∫∫
Ans. I = 2 2
1
9cos +16sindx
x x∫Dividing numerator and denominator
by cos2x
I =
2
2
sec
9 +16tan
xdx
x∫Put tan x = t
∴∴∴∴ sec2 x dx = dt
I = 2
1
16 + 9dt
t∫
I = 2
2
1 1
16 3+
4
dt
t
∫
=( )
–11 1tan +
16 3 4 3 4
tc
=–11 4
tan +12 3
tc
I =–11 4tan
tan +12 3
xc
GROUP (H) – CLASS WORK PROBLEMS
Q-1)1
5+ 4cos x∫∫∫∫
Ans. Let I =5 + 4cos
dt
x∫Put tan
2
x= t,
∴∴∴∴ dx = 2
2
1+
dt
t and cos x =
2
2
1 –
1+
t
t
I =
2
2
2
2
1+
1 –5 – 4
1+
dt
t
t
t
∫
I = ( ) ( )2 2
2
5 1+ – 4 1 –
dt
t t∫
∴∴∴∴ I = 2
2
+ 9
dt
t∫ =( ) ( )
2 22
+ 3
dt
t∫
=–11
2 tan +3 3
tc
=–1
tan2 2tan +3 3
x
c
Q-2)1
3 +3sin 2cosdx
x x++++∫∫∫∫
Ans. Let I = 1
3 + 3sin + 2cosdx
x x∫Put tan =
2
xt
∴∴∴∴ sin x = 2
2
1+
t
t, , , , cos x =
2
2
1 –
1+
t
t
∴∴∴∴ I =
2
2
2 2
2
1+
2 1 –3 + 3 + 2
1+ 1+
dt
t
t t
t t
∫
Mahesh Tutorials Science 37
Indefinite Integration
Q-4)1+ cos cos
cos + cos
xdx
x
αααα
αααα∫∫∫∫
Ans. Let I =αααα
αααα
1+ cos cos
cos + cos
xdx
x∫Put tan =
2
xt ⇒⇒⇒⇒ 2
2=1+
dtdx
t &
2
2
1–cos =
1+
tx
t
I =
2 2α α αα α αα α αα α α
αααα
sin + cos + cos .cos
cos + cos
xdx
x∫
I =( )2 α α αα α αα α αα α α
αααα
sin + cos cos + cos
cos + cos
xdx
x∫
I =2 α αα αα αα α
ααααsin + cos 1.
cos + cos
dxdx
x∫ ∫ ... (i)
Let I1=
ααααcos + cos
dx
x∫ from above sum
I1=
αααα
αααααααα
cot tan1 2 2log
sincot – tan
2 2
x
x
+
∴∴∴∴ I =2
∝∝∝∝
∝∝∝∝∝∝∝∝∝∝∝∝
∝∝∝∝
cot + tan1 2 2sin . log
sincot – tan
2 2
+ .cos +
x
x
x c
∴∴∴∴ I =
∝∝∝∝
∝ ∝∝ ∝∝ ∝∝ ∝∝∝∝∝
cot + tan2 2sin .log + .cos +
cot – tan2 2
x
x cx
Q-5) I =1
3cos2 5++++∫∫∫∫ dxx
Ans. Let I = 1
3cos2 +5dx
x∫Put tanx = t
∴∴∴∴ cos 2x =
2
2
1 –
1+
t
t
Q-3)cos + cos
dx
xαααα∫∫∫∫
Ans. Let I =ααααcos + cos
dx
x∫Put tan =
2
xt ⇒⇒⇒⇒ 2
2=1+
dtdx
t &
2
2
1–cos =
1+
tx
t
I =
( )2
2
2αααα ××××
2
1 –cos + 1+
1+
dt
tt
t
∫
I = ( ) ( )2 2αααα
2
cos 1+ + 1 –
dt
t t∫
I = 2 2α αα αα αα α
2
cos + cos . +1 –
dt
t t∫
I = ( ) ( )2α αα αα αα α
2
1+ cos – 1 – cos
dt
t∫
I =2 2 2α αα αα αα α
2
2cos – 2sin2 2
dt
t
∫
I =2 2 2α αα αα αα α
2
2sin cot –2 2
dt
t
∫
I =2 α αα αα αα α
1 1. .
sin 2cot2 2
αααα
αααα
cot +2
log +
cot –2
t
c
t
I =
αααα
αααααααα
cot + tan1 2 2
log +sin
cot – tan2 2
x
cx
∴∴∴∴ I =
∝∝∝∝
∝∝∝∝∝∝∝∝
cot + tan2 2
cos log +
cot – tan2 2
x
ec cx
=( ) ( )2 2
2
3 1+ + 6 +2 1 –
dt
t t t∫
=( ) ( )
2 22
+ 3 – 2
dt
t∫
=××××
1 +3 – 22 log +
2 2 +3+2
tc
t
=
tan +11 2log +
2tan +5
2
x
cx
Indefinite Integration
38 Mahesh Tutorials Science
Q-6) I =1
2sin2 1++++∫∫∫∫ dxx
Put tan x = t
x = tan–1t 2
1
1dx dt
t
=+
∴∴∴∴ sin 2x =2
2
1
t
t+
∴∴∴∴ I =( )2
2
1 1
2 12 11
dtt t
t
++ +
∫
= 2
1
4 1dt
t t+ +∫
I =( )
2–3
1
2dt
t +∫
=
( ) ( )22
1
2 – 3
dt
t +∫
=1 2 – 3
log2 3 2 3
tc
t
++
+ +
=1 tan 2 – 3
log2 3 tan 2 3
xc
x
++
+ +
GROUP (H) – HOME WORK PROBLEMS
Q-1) 1
3 + 2sinx∫∫∫∫
Ans. I = 1
3+ 2sinx∫Put tan
2
x
= t, then 21sec =1
2 2
x dx
dt
∴∴∴∴ dx =2
2
sec2
dt
x
=2
2
1+ tan2
dt
x
= 2
2
1+
dt
t
∴∴∴∴ I = 2
2
1 2.
2 1+3 + 2
1+
t
t t
t
∫
∴∴∴∴ I = ( )
2
22
1+.1+3 1+ + 4
t dt
tt t∫
∴∴∴∴ I = 2
12
3 + 4 + 3dt
t t∫=
2
2 1
43+ +13
dt
t t∫
∴∴∴∴ I = 2
2 1
3 2 4+ – +13 9
dt
t
∫
∴∴∴∴ I = 2
2 1
3 2 5+ –3 9
dt
t
∫
∴∴∴∴ I = –1
2+
2 1 3tan +3 55
33
t
c
∴∴∴∴ I =–12 3 +2
tan +5 5
tc
∴∴∴∴ I =–1
3tan +22 2tan +
5 5
x
c
dx = 21+
dt
t
∴∴∴∴ I = 22
2
1
1+1 –3 +51+
dt
tt
t
∫
∴∴∴∴ I = ( ) ( )2 23 1 – +5 1+
dt
t t∫
= 22 +8
dt
t∫=
2
1
2 + 4
dt
t∫=
( ) ( )2 2
1
2 + 2
dt
t∫
=–1
××××
1tan +
2 2 2
tc
= –11 tantan +
4 2
xc
Mahesh Tutorials Science 39
Indefinite Integration
Q-4) 1
1+ cos cos xαααα∫∫∫∫
Ans. I = αααα
1
1+ cos cos x∫Put tan
2
x
= t, then 21
sec =12 2
x dx
dt
∴∴∴∴ dx =2
2
sec2
dt
x
=2
2
1+ tan2
dt
x
= 2
2
1+
dt
t
∴∴∴∴ I =22
2αααα
1 2.1+1 –
1+ cos1+
dt
tt
t
∫
∴∴∴∴ I =
2
22 2αααα
1+2 .
1+1+ + cos 1+
t dt
tt t ∫
∴∴∴∴ I =2 2 2α αα αα αα α
12
2sin + 2cos2 2
dt
t
∫
Q-3)1
2+ sin + cosdx
x x∫∫∫∫
Ans. Let I = 1
2+sin + cosdx
x x∫Put tan =
2
xt
∴∴∴∴ sin x = 2
2
1+
t
t, , , , cos x =
2
2
1 –
1+
t
t
Q-2) 1
4 – 5cos x∫∫∫∫
Ans. I = 1
4 – 5cosdxx∫
Put tan 2
x
= t, then 21
sec =12 2
x dx
dt
∴∴∴∴ dx =2
2
sec2
dt
x
=2
2
1+ tan2
dt
x
= 2
2
1+
dt
t
∴∴∴∴ I =( )
1 2.
1 – 1+4 – 5
1+
t
t t
t
∫ 2 2
2
∴∴∴∴ I =( ) ( ) ( )
1+
1+4 1+ – 5 1+
t dt
tt t ∫
2
22 2
∴∴∴∴ I = 2
12
9 –1dt
t∫∴∴∴∴ I =
2
2 1
19–
9
dt
t
∫
∴∴∴∴ I =( )
1–
2 1 3log +119
+233
t
c
t
∴∴∴∴ I =
1tan –
1 2 3log +
13tan +
2 3
x
cx
∴∴∴∴ I =
3tan –11 2log +
33tan +1
2
x
cx
dx = 2
2
1+
dt
t
∴∴∴∴ I = 2 2
2 2
1 2
2 1 – 1+2+ +
1+ 1+
dt
t t t
t t
∫
=( ) ( )2 2
2
2 1+ +2 + 1 –
dt
t t t∫
= 2 + 2 + 3
dt
t t∫
=( ) ( )
22+1 + 2
dt
t∫
=–11 +1
tan +2 2
tc
= –1
+11 2tan +
2 2
xt
c
Indefinite Integration
40 Mahesh Tutorials Science
∴∴∴∴ I =1 1
sin + cos2 2
dt
t
∫2 2 2α αα αα αα α
∴∴∴∴ I =2 2 2α αα αα αα α
1 1
sin + cot2 2
dt
t
∫
∴∴∴∴ I = –1
2 α α αα α αα α αα α α
1 1 1tan
sin cos cos2 2 2
∴∴∴∴ I =–1 αααα
α αα αα αα α
1 1tan tan
2sin cos
2 2
t
∴∴∴∴ I = 2cosec (αααα) tan–1αααα
tan tan +2 2
xc
Q-5)1
5+ 4 sindxx∫∫∫∫
Ans. Let I = 1
5 + 4sindxx∫
Put tan =2
xt
∴∴∴∴ sin x = 2
2
1+
t
t,,,, dx = 2
2
1+
t
t
∴∴∴∴ I =2
2
××××1 2
2 1+5 + 4
1+
dt
t t
t
∫
= ( )2
2
5 1+ + 8
dt
t t∫
= 22
5 +8 +5
dt
t t∫=
2
2
85+ +15
dt
t t∫
=2
2
8 64 645+ + +1 –5 100 100
dt
t t∫
= 2 2
2
5 8 6+ +10 10
dt
t
∫
=–1
××××
8+
2 1 10tan +665
1010
t
c
=–12 5 + 4
tan +5 3
tc
∴∴∴∴ I =–1
5tan + 42 2tan +5 3
x
c
Q-6)1
2 – 3sin2dx
x∫∫∫∫
Ans. Let I =1
2 – 3sin2dx
x∫Put t = tan x tan–1(t) = x
∴∴∴∴ dx = 2
1
1+dt
t
sin 2x = 2
2tan
1+ tan
x
x= 2
2t
1+ t
I =( )2
2
1
1+
22 – 3
1+
dtt
t
t
∫ = ( ) ( )22 1+ – 3 2
dt
t t∫
= 2
1
2 – 6 + 2dt
t t∫ = 2
1 1
2 – 3 +1dt
t t∫=
2
1 1
9 92– 3 + – +1
4 4
dt
t t∫
= 22
1 1
2 3 5– –2 2
t
∫
= ××××
3 5– –
1 1 2 2log +2 3 55
– +22 22
t
c
t
=1 2 – 3 – 5
log +2 5 2 – +3 + 5
tc
t
=( )
( )
2tan – 3 + 51log +
2 5 2tan – 3 – 5
x
c
x
Mahesh Tutorials Science 41
Indefinite Integration
Q-7)1
2 – 3cos2∫∫∫∫ dxx
Ans.Let I =1
2 – 3cos2dx
x∫Put tan = t
∴∴∴∴ dx = 2
1
1+dt
t,
cos 2x =
2
2
1 –
1+
t
t
I =22
2
1
1+1 –2 – 3
1+
dt
tt
t
∫
= ( ) ( )2 22 1+ – 3 1 –
dt
t t∫
= 25 –1
dt
t∫
=( ) ( )
2 25 – 1
dt
t∫
= ××××1 5 –1 1log +
2 5 +1 5c
t
=1 5 tan –1
log +2 5 5 tan +1
xc
x
Q-8)1
3 2sin2 4cos2+ ++ ++ ++ +∫∫∫∫ dxx x
Ans. I =1
3+ 2sin2 + 4cos2dx
x x∫Put tan x = t x = tan–1 t
∴∴∴∴ sin x = 2
2
1+
t
t
cos x =
2
2
1 –
1
t
t+
dx = 21+
dt
t
I =( )2 2
2 2
1
2 1 – 1+3 +2 + 4
1+ 1+
dt
t t t
t t
×
∫
= 2 2
1
3+ 3 + 4 + 4 – 4dt
t t t∫
Q-9)1
3sin 4 cos 5+ ++ ++ ++ +∫∫∫∫ dxx x
Ans. I =1
3sin 4cos 5+ +∫ dxx x
Put tan 2
x = t
∴∴∴∴ x = 2tan–1 t
dx = 2
2
1+dt
t
∴∴∴∴ sin x = 2
2
1+
t
t,
cos x =
2
2
1 –
1+
t
t
I = 22
2 2
1 2
1+2 1 –3 + 4 +51+ 1+
dt
tt t
t t
∫
= 2 2
12
6 + 4 – 4 +5 +5dt
t t t∫
= 2 2
12
6 + 4 – 4 +5 +5dt
t t t∫= 2
12
+ 6 – 9dt
t t∫= ( )
–22 +3t dt∫ =
–12 +
+3c
t
I =–2
+
tan + 32
cx
= 2
1
7 + 4 –dt
t t∫= 2
1
11 – 4 – 4 –dt
t t∫
=( ) ( )
2 2
1
11 – – 2dt
t∫
=( )
( )
11+ – 22log +
2 11 11 – – 2
tc
t
=2 11 + tan – 2
log +2 11 11 – tan +2
xc
x
I =1 tan + 11 – 2
log +2 11 11 +2 – tan
xc
x
Indefinite Integration
42 Mahesh Tutorials Science
Q-10)1
3 4 sin++++∫∫∫∫ dxx
Ans. I =1
3 4sin+∫ dxx
Put tan 2
x = t
∴∴∴∴ x = 2tan–1 t
∴∴∴∴ sin x = 2
2
1+dt
t
dx = 2
2
1+
t
tdt
I =( )2
2
1 2
2 1+3+ 41+
dt
t t
t
∫
= ( )2
12
3 1+ +8dt
t t∫
= 2
12
3+3 +8dt
t t∫=
2
2 1
83+ +13
dt
t t∫
=2
2 1
4 16 73+ 2 + –
3 9 9
dt
t t
∫
=2
2 1
3 4 7+ –3 9
dt
t
∫
= 22
2 1
3 4 7+ –3 3
dt
t
∫
=
4 7–
2 1 3 3log3 4 772
3 33
t
t
+
+ +
=
3tan 4 – 71 2log7 3tan 4 7
2
+
+
+ +
x
cx
GROUP (I) – CLASS WORK PROBLEMS
Q-1) 1
3 sin + 4cosdx
x x∫∫∫∫
Ans. Let I = 1
3sin + 4cosdx
x x∫
=1 1
3 45sin + cos
5 5
dx
x x∫
=θ θθ θθ θθ θ
1
45cos sin + sin cos
5
dx
x x∫
Where cos θθθθ =3
5 and
sin θθθθ =4
5
∴∴∴∴ tan θθθθ =4
3
∴∴∴∴ I =( )θθθθ
1
5 sin +
dx
x∫
∴∴∴∴ θθθθ = tan–14
3
= ( )θθθθ1
cos +5
ec x dx∫
=θθθθ1
log tan + +5 2 2
xc
=
–1 4+ tan
1 3log tan +
5 2
x
c
Q-2)1
cos + 3 sindx
x x∫∫∫∫
Ans. Let I =1
1.cos + 3 sindx
x x∫Let 3 = r cos αααα and 1 = r sin αααα
∴∴∴∴2
3 + 12 = r2 and tan α α α α = 1
3
∴∴∴∴ 2 = r and α α α α = ππππ
6
3 sinx + 1 cosx = r cosαααα sinx + r sin αααα cosx
Mahesh Tutorials Science 43
Indefinite Integration
Q-3)1
1+ sindx
x∫∫∫∫
Ans. Let I =1
1+sindx
x∫=
1
cos + sin2 2
dxx x∫
=1
1 12 cos + sin2 22 2
dx
x x∫
=π ππ ππ ππ π
1
2 sin .cos + cos sin4 2 4 2
dx
x x∫
=ππππ
1
2sin +
2 4
dx
x
∫
=ππππ1
cos +2 42
xec dx
∫
=
ππππ
××××
+1 12 4log tan +
122
2
x
c
=ππππ1
log tan + +4 82
xc
GROUP (I) – HOME WORK PROBLEMS
Q-1) 1
2sin – cosdx
x x∫∫∫∫
Ans. I = 1
2sin – cosdx
x x∫2 = r cos ∝∝∝∝ and 1 = r sin ∝∝∝∝
∴∴∴∴ (2)2 + (1)2 + r2 and tan ∝ ∝ ∝ ∝ = 2
∴∴∴∴ r = 5 , ∝ ∝ ∝ ∝ = tan–1(2)
∴∴∴∴ I =2sin –1.cos
dx
x x
=∝ ∝∝ ∝∝ ∝∝ ∝5 cos .sin – 5 sin .cos
dx
x x∫= ( )∝∝∝∝5 sin –
dx
x∫
= ( )∝∝∝∝1
cos –5
ec x dx∫=
∝∝∝∝1 –log tan +
25
xc
= ( )–1– tan 21log tan +
25
xc
Q-2)1
3sin – 4cosdx
x x∫∫∫∫
Ans. Let I =1
3sin – 4cosdx
x x∫= 1
3 45sin – cos
5 5
dx
x x∫
Let, 3
5 = cos a and
4
5 = sina
∴∴∴∴ tan a = αααα
αααα
sin 4=
cos 5
∴∴∴∴ a = tan–1 4
3
So,
I =α αα αα αα α
1 1
5 cos sin – sin cosdx
x x∫
= ( )αααα
1 1
5 sin –dx
x∫
∴∴∴∴ 3 sinx + cosx = r sin (x + αααα)
∴∴∴∴ I = ( )ααααsin +
dx
r x∫
∴∴∴∴ I = ( )αααα1
cos +ec x dxr ∫
∴∴∴∴ I =αααα1 +
log tan +2
xc
r
∴∴∴∴ I =
ππππ+
1 6log tan +2 2
x
c
Indefinite Integration
44 Mahesh Tutorials Science
= ( )αααα1
cos –5
ec x dx∫
=αααα1 –
log tan +5 2
xc
=
–1 4– tan
1 3log tan +
5 2
x
c
Q-3)1
sin + 3 cosdx
x x∫∫∫∫Ans. Let, I =
1
sin + 3 cosdx
x x∫=
1 1
2 1 3sin + cos
2 2
dx
x x∫
=π ππ ππ ππ π
1 1
2cos sin + sin cos
3 3
dx
x x∫
=ππππ
1 1
2sin +
3
dx
x
∫
=ππππ1
cos +2 3
ec x dx ∫
=ππππ1
log tan + +2 2 6
xc
Q-4)1
1+ sin2dx
x∫∫∫∫Ans. Let, I =
1
1+sin2dx
x∫
=( )
2
1
cos + sin
dx
x x∫
=cos +sin
dx
x x∫
=
1
1 12 cos + sin22 2
dx
xx
∫
=π ππ ππ ππ π
1
2 sin .cos + cos sin4 2 4
dx
xx
∫
=ππππ
1
2 sin +4
dx
x
∫
=ππππ1
cos +42
ec x dx ∫
=
ππππ+
1 4log tan +22
x
c
=ππππ1
log tan + +2 82
xc
Q-5)1
5cos 12sindx
x x++++∫∫∫∫Ans. I =
+∫1
5cos 12sindx
x x
Put tan2
x = t
∴∴∴∴ x = 2 tan–1t
dx = 2
2
1+
dt
t, sin x = 2
2
1+
dt
t, cos x =
2
2
1 –
1+
t
t
I =( )2 2
2 2
1 2
1 – 2 1+5 +121+ 1+
dt
t t t
t t
∫
= 2
12
5 – 5 +24dt
t t∫=
2
2 1
2451 – +
5
dtt
t∫
=2
2 1
5 169 12 144– – 2 +
25 5 25
dt
t t
∫
= 2 2
2
2 1
5 13 12– –
5 5
dt
t
∫
=
+
+
+
∫13 12
–2 1 5 5log
13 12135–2
5 55
t
c
t
I =
1+ tan
1 5 2log +13
5 – tan2
x
cx
Mahesh Tutorials Science 45
Indefinite Integration
GROUP (J) – CLASS WORK PROBLEMS
Q-1)
2
2 2–
xdx
a x∫∫∫∫
Ans. I =–
xdx
a x∫2
2 2
Put x = a sin θθθθ
∴∴∴∴ dx = a cos θ θ θ θ d θθθθ
∴∴∴∴ I =
2 2
2 2 2
θ θ θθ θ θθ θ θθ θ θ
θθθθ
sin . cos
– sin
a a d
a a∫
=2 2
θ θθ θθ θθ θsina d∫=
2θθθθ
1 – cos2
2a d∫
=
2
θ θθ θθ θθ θ1
– sin2 +2 2
ac
= [ ]2
θ θ θθ θ θθ θ θθ θ θ– sin cos +2
ac
We resubstitute for sin θ θ θ θ and cos θ θ θ θ as follows.
∵∵∵∵ sin θ θ θ θ = x
a, we draw a triangle with one angle
900 and another marked as θθθθ. The opposite
side of θθθθ is taken as x and hypotenuse as a,
giving sin θθθθ = x
a. The remaining side
is 2 2–a x .
∴∴∴∴ cos θ θ θ θ = 2 2–a x
a.
∴∴∴∴ I =
2 2 2–1 –
sin – . +a x x a x
ca a a a
x a
θ
a x2 2 –
Q-2)–
xdx
a x∫∫∫∫
Ans. Let I =–
xdx
a x∫Put x = a sin2 θθθθ
∴∴∴∴ dx = 2a sin θ θ θ θ cos θθθθ dθθθθ
∴∴∴∴ I =
2
2
θθθθ
θθθθ
sin
– sin
a
a a∫ .(2a sin θθθθ. cos θθθθ dθθθθ)
=2
θ θθ θθ θθ θ2 sina d∫
=θθθθ
θθθθ1 – cos2
22
a d ∫
=θθθθ
θθθθsin2
– +2
a c
= a(θ θ θ θ – sin θθθθ. cos θθθθ)+c
=–1 . –
sin – +x x a x
a ca a
= ( )–1sin – – +x
a x a x ca
x
a x –
θ
a
Q-3) I =–
++++∫∫∫∫a x
dxa x
Put x = a cos 2θθθθ ⇒⇒⇒⇒ cos 2θθθθ
=–11
cos2
x x
a aθ
⇒ =
∴∴∴∴ dx = – 2a sin 2θθθθ dθθθθ
∴∴∴∴ I = – cos2
– 2 sin2cos2
a aa d
a a
θθ θ
θ+∫
=1 – cos2
–2 .sin21 cos2
a dθ
θ θθ+∫
=sin
–2 .2sin coscos
a dθ
θ θ θθ∫
Indefinite Integration
46 Mahesh Tutorials Science
Q-5) 2 2
1
–1x x
Ans. Let I = 2 2
1
–1dx
x x∫Put x = sec θθθθ
∴∴∴∴ dx = sec θ θ θ θ tan θ θ θ θ dθθθθ
∴∴∴∴ I = 2 2θ θθ θθ θθ θ
1
sec sec –1∫ sec θθθθ tan θθθθ dθθθθ
= 2
θ θθ θθ θθ θθθθθ
θ θθ θθ θθ θ
sec tan
sec tand∫
=1
secd∫ θθθθ
θθθθ
= θ θθ θθ θθ θcos d∫= sinθθθθ + c .... (i)
x
1
θ
x 2 – 1
Now, sec θθθθ = x
∴∴∴∴ cos θθθθ =1
x
∴∴∴∴ sin θθθθ =–1x
x
2
....[from figure]
From (i), we get
I =–1
+x
cx
2
Q-4)(((( ))))
22 2
1
+dx
x a∫∫∫∫
Ans. Let I =( )2 2
1
+dx
x a∫Put x = a tan θθθθ
∴∴∴∴ dx = a sec2 θ θ θ θ dθθθθ
∴∴∴∴ I =( )
22 2 2
θθθθ
1
+ tan +a a∫ .(a sec2 θθθθ dθθθθ)
=
2
4 4
θ θθ θθ θθ θ
θθθθ
sec
sec
a d
a∫
=2
3θ θθ θθ θθ θ
1cos d
a ∫
=
2
3
θθθθθθθθ
1 1+ cos
2d
a
∫
=
2
3
θθθθθθθθ
1 sin+ +
22c
a
∫
= ( )3
θ θ θθ θ θθ θ θθ θ θ1
+sin .cos +2
ca ∫
x
a
θ
x a2 2 +
=2–2 2sina dθ θ∫
= ( )–2 1 – cos2a dθ θ∫=
sin2–2 –
2a
θθ
=–1 21 1
–2 cos + 1 – cos 2 +2 2
xa c
aθ
=
2–1
2– cos + 1 – +
x xa a c
a a
=–1 2 2– cos + – +x
a a x ca
I =–1
3 2 2 2 2
1tan + .
2 + +
x x a
aa x a x a
I =–1
3 2 2
1tan + +
2 +
x axc
aa x a
Mahesh Tutorials Science 47
Indefinite Integration
GROUP (J) – HOME WORK PROBLEMS
Q-1)2 2
2
–a xdx
x
Ans. I =2 2
2
–a xdx
x
Put x = a sin θθθθ
∴∴∴∴ dx = a cos θ θ θ θ dθθθθ
∴∴∴∴ I =
2 2 2
2 2
θθθθ
θθθθ
– sin
sin
a a
a∫ a cos θθθθ dθθθθ
= 2 2
θ × θ θθ × θ θθ × θ θθ × θ θ
θθθθ
cos cos
sin
a a d
a∫
=
2
2
θθθθθθθθ
θθθθ
cos
sind∫
=2
θ θθ θθ θθ θcot d∫
= ( )2θ θθ θθ θθ θcos –1ec d∫
=2θ θ θθ θ θθ θ θθ θ θcos –ec d d∫ ∫
= –cot θθθθ – θ θ θ θ + c ....(i)
a x
θ
a 2 – x
2
Now, x = a sin θθθθ
∴∴∴∴ sin θθθθ =x
a
∴∴∴∴ θ θ θ θ = sin–1x
a
Also, cos θθθθ =–a x
a
2 2
....[from figure]
∴∴∴∴ cot θθθθ =θθθθ
θθθθ
cos
sin
=
–a x
ax
a
2 2
=2 2–a x
x
From (i), we get
I =2 2
–1–– sin +
a x xc
x a
Q-2)1 –
1+
xdx
x∫∫∫∫
Ans. Let I =1 –
1+
xdxx∫
Put x = cos θθθθ
∴∴∴∴ dx = –sin θ θ θ θ dθθθθ
∴∴∴∴ I =θθθθ
θθθθ
1 – cos
1+ cos∫ (–sin θθθθ dθθθθ)
=
2
2
θθθθ
θ θθ θθ θθ θθθθθ
θθθθ
2sin2 –2sin .cos
2 22sin
2
d ∫
=
θθθθ
θ θθ θθ θθ θθθθθ
θθθθ
sin2 –2sin .cos
2 2cos
2
d ∫
=2 θθθθ
θθθθ–2 sin2d∫
=θθθθ
θθθθ1 – cos
–22
d ∫
= ( )θ θθ θθ θθ θcos –1 d∫= sin θθθθ – θ θ θ θ + c
1
x
θ
1 – x2
∴∴∴∴ I = 21 – x – cos–1(x) + c
Indefinite Integration
48 Mahesh Tutorials Science
Q-4)–a x
dxx
Ans. I =–a x
dxx
Put x = a sin2 θθθθ
∴∴∴∴ dx = 2a sin θ θ θ θ cos θ θ θ θ dθθθθ
∴∴∴∴ I =
2
2
θθθθ
θθθθ
– sin
sin
a a
a∫ 2a sin θθθθ cos θ θ θ θ dθθθθ
=
2
2
θθθθ
θθθθ
cos
sin
a
a∫ . 2a sin θθθθ cos θ θ θ θ dθθθθ
=2
θ θθ θθ θθ θ2 cosa d∫=
θθθθθθθθ
1+cos22
2a d∫
θ
ax
a x–
=θθθθ
θθθθsin2
. + +2
a c
= a(θθθθ + sin θ θ θ θ cos θθθθ) + c
=–1 . –
sin + +x x a x
a ca a
= ( )–1sin + – +x
a x a x ca
GROUP (K) – CLASS WORK PROBLEMS
Q-1) cosx x dx∫∫∫∫
Ans. Let I = .cosx x dx∫We know that,
uvdx∫ = – .du
u v dx v dx dxdx
∫ ∫
Q-3)
(((( ))))3
2 2
1
+a x∫∫∫∫
Ans. Let I =
( )3
2 2
1
+
dx
a x∫
Put x = a tan θθθθ
∴∴∴∴ dx = a sec2 θ θ θ θ dθθθθ
∴∴∴∴ I = ( )
2
32 2 2
θθθθθθθθ
θθθθ
sec
+ tan
ad
a a∫
=
( )
2
32 2
θθθθθθθθ
θθθθ
sec
1+ tan
ad
a
∫
=
2
32 2
θθθθθθθθ
θθθθ
sec
sec
ad
a ∫
=
2
6 6
θθθθθθθθ
θθθθ
sec
sec
ad
a∫ =
2
3 3
θθθθ
θθθθ
sec
sec
a
a∫
= 2
θθθθ
θθθθ
1
sec
d
a ∫ = 2θ θθ θθ θθ θ
1cos d
a ∫= 2
θθθθ1sin + c
a.... (i)
a
x
θ
x 2 – a
2
Now, x = a tan θθθθ
∴∴∴∴ tan θθθθ =x
a
∴∴∴∴ sin θθθθ = 2 2+
x
x a....[from figure]
From (i), we get
I =2 2 2
1. +
+
xc
a x a
=2 2 2
++
xc
a x a
Mahesh Tutorials Science 49
Indefinite Integration
Q-2)x
xe dx∫∫∫∫
Ans. Let I =xxe dx∫
= ( ) ( ) 1x x– +
dx e dx e dx x dx c
dx
∫ ∫
= 1x x– .1 +xe e dx c∫
= xex – ex + c
Q-3)3 logx xdx∫∫∫∫
Ans. Let I =3 logx xdx∫
∴∴∴∴ I = ( ) 3log x x dx∫
∴∴∴∴ I =
( )
3
3
log
– log
x x dx
dx x dx dx
dx
∫∫ ∫
∴∴∴∴ I =
4 41log – .
4 4
x xx dx
x
∫
∴∴∴∴ I =
43log 1
–4 4
x xx dx∫
∴∴∴∴ I =4 4log 1
– +4 4 4
x x xc
∴∴∴∴ I = ( )4
4log –1 +16
xx c
Q-4)1+ sin
xdx
x∫∫∫∫
Ans. Let I =1+sin
xdx
x∫I = ××××
1– sin
1+sin 1– sin
x xdx
x x∫I = 2
– sin
1– sin
x x xdx
x∫
I = 2
– sin
cos
x x xdx
x∫I =
2sec – sec tanx x x x dx∫uv dx∫ = –
duv dx v dx dx
dx
∫ ∫ ∫
∴∴∴∴ I = tan – 1.tan
– sec – 1.sec
x x x dx
x x x dx
∫∫
= x tan x – log |sec x|
– [x sec x – log |sec x + tan x|] + c
= x tan x – log |sec x|
+ log|sec x + tan x| + c
= x (tan x – sec x)
+ log sec + tan
+sec
x xc
x
= x (tan x – sec x) + log|1 + sin x| + c
.cosx x dx∫ =
( ) 1
. cos
– cos +
x x dx
dx dx x dx c
dx
∫∫ ∫
∴∴∴∴ x. sin x – 1sin .1 +x dx c∫∴∴∴∴ x sin x + cos x + c
Q-5)2cosx x dx∫∫∫∫
Ans. Let I =2cosx x dx∫
=2 2cos – cos .
dx x dx x dx x dx
dx
∫ ∫ ∫
... (i)
∴∴∴∴2cos x dx∫
=1+cos2
2
xdx∫
=1 sin2
+2 2
xx
and 1 sin2
+2 2
xx dx ∫
Indefinite Integration
50 Mahesh Tutorials Science
Q-6)2. xx e dx∫∫∫∫
Ans. Let I =2 x.x e dx∫
2 x.x e dx∫
= ( ) ( )2 2
1
x x– +d
x e dx e dx x dx cdx
∫ ∫
= ( )2
1
x x. – . 2 +x e e x dx c∫=
2
1
x x. – 2 . +x e x e dx c∫
∴∴∴∴ I = ( )2
2
x x x– 2 – .1 +x e x e e dx c ∫
= x2ex – 2xex + 2ex + c.
Q-7)–1. tanx x dx∫∫∫∫
Ans. Let I = ( )–1tanx x dx∫
= ( )–1tan .x x dx∫
∴∴∴∴ =
2 2
12
1tan . – . +
2 2 1+
x xx dx c
x
∫–1
=( )2
2
12
+1 –11 1tan – +
2 2 1+
xx x dx c
x∫–1
=2
12
1 1 1tan – 1 – +
2 2 1+x x dx c
x
∫–1
=2 –11 1tan – – tan + .
2 2x x x x c
–1
Q-8)–1tan x dx∫∫∫∫
Ans. Let I =–1tan x dx∫
dx = ( ) ( )–1tan . 1x dx∫
= ( )–1tan 1 – tan 1d
x dx x dx dxdx
∫ ∫ ∫–1
= ( ) ( )–1
2
1tan – .
1+x x x dx
x∫
= –1
2
1 2tan –
2 1+
xx x dx
x∫Now,
d
dx(1 + x2) = 2x and
( )
( )
'f x
f x∫ = log|f(x) + c
∴∴∴∴ I = x tan–1x – 1
2 log(1 + x2)+c
( )21+ > 0x∵
Q-9) (((( ))))2
log x dx∫∫∫∫Ans. = (log x)2 ( )
21 – log 1
ddx x dx dx
dx
∫ ∫ ∫
= log x2.x – × ×× ×× ×× ×1
2log x dxx∫
= x (log x)2 – ( )2log 1x dx∫= x (log x)2 – 2 I
1
I1= ( ) ( )log 1x dx∫= ( )log 1 – log 1
dx dx x dx dx
dx
∫ ∫ ∫
= ( ) ××××1
log –x x xdxx∫
= x log x – x + c
= x (log x – 1) + c
I = x (log x)2 – 2x (log x – 1) + c
=
21 1+ cos2
2 2 8
xx
∴∴∴∴ I =1 sin2 1 sin2. + – +2 2 2 2
x xx x x dx
∫
∴∴∴∴ I =
2 2.sin2 1+ – + cos2 +
2 4 4 8
x x x xx c
∴∴∴∴ I =
2 sin2 cos2+ + +
4 4 8
x x x xc
Mahesh Tutorials Science 51
Indefinite Integration
Q-10) (((( ))))2 2log + +∫∫∫∫ x x a dx
Ans. I = ( )2 2log + +x x a dx∫I = ( )2 21.log + +x x a dx∫= ( )
( )
2 2
2 2
log + +
1 – 1 . log + + .
x x a
ddx dx x x a dx
dx
∫ ∫ ∫
= ( )2 2
2 2 2 2
. log + +
2– 1+
+ + 2 +
x x x a
x xdx
x x a x a
∫
= ( )2 2
2 2. log + + –
+
xx x x a dx
x a∫In the second integral
Put x2 + a2 = t
∴∴∴∴ x dx =2
dt
∴∴∴∴ I = ( )2 2
2 2
1 2.log + + –
2 +
xx x x a dx
x a∫= ( ) ( )2 2 2 21
.log + + – 2 + +2
x x x a x a c
= ( )2 2 2 2. log + + – + +x x x a x a c
Q-11) cos xdx∫∫∫∫Ans. Let I = cos xdx∫
Put x = t
Then x = t2 and dx = 2t dt
I = ( )cos .2t t dt∫I = 2 cost tdt∫I = ( )2 cos – cos
dt tdt t tdt dt
dt
∫ ∫ ∫
I = 2 sin – sint t tdt ∫
I = 2(t sin t + cos t) + c
I = 2t sin t + 2 cos t + c
I = 2 sin + 2cos +x x x c
Q-12) (((( ))))sin . log cosx x dx∫∫∫∫Ans. Let I = ( )sin .log cosx x dx∫
Put cos x = t ⇒⇒⇒⇒ sin x dx = – dt
I = ( )log cos sinx x dx∫I = ( )log t dt−∫I = ( )– log .t dt∫I = ( ) ( )– log 1. – log 1
dt dt t dt dt
dt
∫ ∫ ∫
I = ( ) ( )1
– log –t t t dtt
∫I = – log – 1t t dt
∫
I = –t log t + t + c
I = t (1 – log t) + c
I = cos x {1 – log (cos x)} + c
Q-13)23 x
x e dx∫∫∫∫Ans. I =
23 xx e dx∫=
22 x.x x e dx∫Put x2 = t 2x dx = dt
I =t1
2t e dt∫
=t t1
–2
dtt e dt e dt
dt
∫ ∫ ∫
=t t1–
2t e e dt ∫
=t t1– +
2t e e c
= [ ]t1–1 +
2e t c
=
22
x
x –1 +2
ee c
Indefinite Integration
52 Mahesh Tutorials Science
Q-15) . sin∫∫∫∫ xe x dx
Ans. Let I =x .sine x dx∫
=x xsin – . sin
dx e dx e dx x dx
dx
∫ ∫
=x xsin . – cosx e e x dx∫
=
( )
x
x x
sin .
– cos – – sin
x e
x e dx e x dx ∫ ∫
∴∴∴∴ I =x x x.sin – cos. – sine x e e dx∫
∴∴∴∴ I = ex(sin x – cos x) – I
∴∴∴∴ 2I = ex(sin x – cos x)
∴∴∴∴ I =
x
2
e(sin x – cos x) + c
Q-16) cos (2 log ) x dx∫∫∫∫Ans. Put 2logx = t
∴∴∴∴ log x2 = t
∴∴∴∴ 2
t
e = x
2
x dx = dt
dx =2
dt · 2
t
e
I = 2
t1
cos ·2
t e dt∫
I =2 2
t t1
cos – (cos ) 2
dt e dt e dt t dt
dt
∫ ∫∫
= ( )2 21
t t1
2cos · – 2 · – sin +2
t e e t dt c ∫
= ( )2 2 2
t t t
cos + sin – sind
t e t e dt e dt tdt
∫ ∫
= cos t · 2
t
e + 2 sint 2
t
e
2
t
–2 cos +e t dt c∫ I = cos t · 2
t
e + 2 sint 2
t
e – 4I + c
I =5
x [cos(2 log x) + 2 sin (2 log x)] + c
Q-14)3sec∫∫∫∫ x dx
Ans. Let I =3sec x dx∫
I = ( )2sec secx x dx∫
=
( )
2
2
sec sec
– sec sec
x x dx
dx dx dx
dx
∫∫ ∫
= sec x tan x
– 1sec tan .tan +x x x dx c∫= sec x tan x
( )2
1– sec sec –1 +x x dx c∫= sec x tan x
( )2
1– sec sec –1 +x x dx c∫= sec x tan x
3
1– sec + sec +x dx x dx c∫ ∫2I = sec x tan x + log|sec x + tan x| + c
2
I = 1
2[sec x tan x + log|sec x + tan x|+ c
GROUP (K) – HOME WORK PROBLEMS
Q-1)2secx x dx∫∫∫∫
Ans. =2 2sec – sec ( )
dx x dx x dx x dx
dx
∫ ∫ ∫
= 1tan – tan · 1 +x x x dx c∫= x tanx – log |secx| + c
= x tanx + log|cosx| + c
Mahesh Tutorials Science 53
Indefinite Integration
Q-2) sinx x dx∫∫∫∫Ans. Let = sinx x dx∫
= sin – sin ·d
x x dx x dx x dxdx
∫ ∫ ∫
= x (– cos x) – – cos .1x dx∫= –x cos x + sin x + c
Q-3)2 logx x dx∫∫∫∫
Ans. Let =2 logx x dx∫
Here neither an exponential nor a
trigonometric function is present we take
the algebric function x2 as
∴∴∴∴ v dx∫ =2 log .x x dx∫
∴∴∴∴ ( )2log .x x dx∫=
3 3 1log – . +
3 3
x xx dx c
x
∫
=3 2
1
1 1log – +
3 3x x x dx c∫
=3 3
1
1 1log – +
3 9x x x c
Q-4)2.cosx x dx∫∫∫∫
Ans. Let I =2.cosx x dx∫
= cos – cosd
x x dx x dx x dxdx
∫ ∫ ∫2 2
= ( ).sin – sin 2x x x x dx∫2
= 2.sin – 2
sin – sin
x x
dx x dx x dx x dx
dx
∫ ∫ ∫
= x2. sin x – 2 – cos + cos .1x x x dx ∫
= x2. sin x + 2 x cos x – 2 sin x + c
Q-5)1 – cos2
xdx
x∫∫∫∫Ans. Let I =
1– cos2
xdx
x∫= 22sin
xdx
x∫=
21.cos
2x ec x dx∫
=
( )
2
2
cos1
2– cos .
x ec x dx
dec x dx x dx
dx
∫∫
= ( ) ( )1
– cot – – cot 12x x x dx ∫
=1
2[–x . cot x + log |sin x|] + c.
Q-6)2 –1cosx x dx∫∫∫∫
Ans. Let I =2 –1cosx x dx∫
= ( )–1 2cos .x x dx∫= ( )
( )
–1
2 –1 2
cos
– cos .
x
dx dx x x dx dx
dx
∫ ∫ ∫
= ( )3 3
–1
2
–1cos – .
3 31+
x xx dx
x∫
=
3 –1 3
2
cos 1+
3 3 1+
x x xdx
x∫
=
3 –1 2
2
cos 1+ .
3 3 1 –
x x xx dx
x∫Put 1 – x2 = t
∴∴∴∴ –2x dx = dt
∴∴∴∴ x dx =–
2
dtand x2 = 1 – t
∴∴∴∴ I =( )3 –1 –cos 1 1 –
+ .3 3 2
dtx x t
t∫=
3 –1cos 1 1 ––
3 6
x x tdt
t∫
Indefinite Integration
54 Mahesh Tutorials Science
Q-7)–1sin xdx∫∫∫∫
Ans. =–1sin .1x dx∫= –
duuvdx u vdx vdx dx
dx
∫ ∫ ∫ ∫
∴∴∴∴ I = sin–1 x.x – 2
+ ,1 –
xdx c
x∫∴∴∴∴ 1 – x2 = t ∴∴∴∴ – 2x dx = dt
∴∴∴∴ x dx = –
2
dt
I = x sin–1 x – 1 –1
+2dt c
t
∫
= x sin–1 x + ( )12 +
2t c
= x sin–1x+ 21 – +x c
Q-8) sin xdx∫∫∫∫Ans. =x t
∴∴∴∴ x = t 2
∴∴∴∴ dx = 2t dt
I = sin .2t tdt∫ = 2 sint tdt∫= ( ) ( )2 cos – cos .1 +t t t dt c
∫
= 2 [– t cost + sin t] + c
= –2 cos + sin +x x x c
Q-9)–1tan x dx∫∫∫∫
Ans. Let I =–1tan x dx∫
Put = t ∴∴∴∴ x = t2 ∴∴∴∴ dx = 2t dt
∴∴∴∴ I = ( ) ( )–1tan 2t t dt∫= ( )
( )
–1
–1
tan . 2
– tan . 2
t t dt
dt t dt dt
dt
∫∫ ∫
= t2 tan–1 t –2
2
1.
1+t dt
t∫= t2 tan–1 t –
( )2
2
1+ –1
1+
tdt
t∫= t2 tan–1 t – 2
1 +1+
dtdt
t∫ ∫= t2 . tan–1 t – t + tan–1 t + c
= (t2 + 1)tan–1 t – t + c
= (x + 1)tan–1 – + .x x c
Q-10) (((( ))))cos . log sinx x dx∫∫∫∫Ans. Let I = ( )cos .log sinx x dx∫
Put sin x = t
∴∴∴∴ cos x dx = dt
∴∴∴∴ I = log t dt∫= ( )log .1t dt∫= log 1 – 1 log
dt dt dt t dt
dt
∫ ∫ ∫= t. log t
1– t dt
t
∫
= t. log t – dt∫ = t. log t – t + c
= t (log t – 1) + c = sin x [log (sin x)–1] + c
Q-11) .cosxe x dx∫∫∫∫
Ans. Let I =x .cose x dx∫
=
3 –1cos 1 1– –
3 6
x xt dt
t
∫
=
1 13 –1–2 2
cos 1 1– +
3 6 6
x xt dt t dt∫ ∫
=
1 3–
3 –1 2 2cos 1 1– . + +
1 33 6 6
2 2
x x t tc
= ( ) ( )1 33
–1 2 22 21 1
cos – 1 – + 1 – +3 3 9
xx x x c
=( )
( )
32 2 13
–1 2 21 –1
cos + – 1 – +3 3 3
xxx x c
Mahesh Tutorials Science 55
Indefinite Integration
Q-12) (((( ))))sin log x dx∫∫∫∫Ans. Let I = ( )sin log x dx∫
Put log x = t
∴∴∴∴ x = et
∴∴∴∴ dx = et dt
∴∴∴∴ I =tsin .t e dt∫
=t tsin . – sin
dt e dt e dt t dt
dt
∫ ∫ ∫= et. sin t t– cose t dt∫= ( )t t t.sin – cos – – sine t t e dt e t dt
∫ ∫
∴∴∴∴ I = et. sin t – cos t. et – I + c1
∴∴∴∴ 2I = et (sin t – cos t) + c1
∴∴∴∴ I =
t
2
e(sin t – cos t) +
1
2
c
∴∴∴∴ I =
t
2
e(sin t – cos t) + c, Where c =
1
2
c
Q-13)3cosec x dx∫∫∫∫
Ans. Let I =3cosec x dx∫
=2cos .cosec x ec x dx∫
Q-14) ∫∫∫∫ 3sec 4x dx
Ans. Let I = sec 4∫ 3x dx
=
( )
sec 4 sec 4
– sec 4 sec4
∫
∫ ∫
2
2
x x dx
dx dx x dx
dx
=
( )
tan4sec 4 .
4
tan4– .sec 4 .tan4 4 .
4
∫
xx
xx x dx
= ( )sec4 .tan4
– sec 4 –1 .sec4 .4 ∫ 2x x
x x dx
=sec 4 .tan4
– sec 4 + sec 44 ∫ ∫3x x
x dx x dx
I =sec4 .tan4
–4
1+ log sec4 + tan4 +
4×××× 1
x xI
x x c
2I =log sec4 + tan4sec 4 .tan4
+ +4 4
1
x xx xc
∴∴∴∴ I = 1
8[sec 4x.tan 4x + log|sec 4x + tan 4x|] + c
=
( )
2
2
cos cos
– cos . cos
ec x ec x dx
dec x ec x dx dx
dx
∫ ∫∫ ∫
= (cosec x)(– cot x)
( ) ( )– – cos .cot – cotec x x x dx∫= – cosec x cot x
2– cos .cotec x xdx∫= – cosec x cot x ( )2– cos cos –1ec x ec x dx∫= – cosec x cot x
3– cos + cosec dx ec x dx∫ ∫∴∴∴∴ I = – cosec x cot x – I + log|cosec x – cot x|
∴∴∴∴ 2I = – cosec x cot x + log|cosec x – cot x|
∴∴∴∴ I =– cos cot 1
+2 2
ec x x
log|cosec x – cot x| + c
or I =– cos .cot 1
+ log tan +2 2 2
ec x x xc
=x xcos – . cos
dx e dx e dx x dx
dx
∫ ∫
= ( )x xcos . – – sinx e e x dx∫= x
x x
cos .
+ sin – cos
x e
x e dx e x dx ∫ ∫
∴∴∴∴ I = cos.ex + sin x.ex – I + c1
∴∴∴∴ 2I = ex(sin x + cos x) + c1
∴∴∴∴ I =x
2
e(sin x – cos x) +
1
2
c
∴∴∴∴ I =x
2
e(sin x – cos x) + c, where c =
1
2
c
Indefinite Integration
56 Mahesh Tutorials Science
GROUP (L) – CLASS WORK PROBLEMS
Q-1) (((( ))))2cot – cot – 1xe x x dx∫∫∫∫
Ans. Let I = ( )2x cot – cot –1e x x dx∫= ( )2x cot – cose x ec x dx∫
Put f (x) = cot x. Then
f ′′′′(x) = – cosec2 x
∴∴∴∴ I = ( ) ( )'x +e f x f x dx ∫= ex f(x) + c
= ex cot x + c
Q-2)(((( ))))
2
1+
2 +
x xe dx
x∫∫∫∫
Ans. Let I = ( )2
x 1+
2+
xe dx
x∫
I =( )
( )2
x2+ –1
2+
xe dx
x∫
I =( )
2
x 1 1–
2+ 2+e dx
x x
∫
I =( )
( )2
x–11
+2+ 2+
e dxx x
∫
Put f(x) =1
2+ x
Then f ′′′′(x) = ( )2
–1
2+ x
So,
I = ( ) ( )'x +e f x f x dx ∫∴∴∴∴ I = ex f(x) + c
∴∴∴∴ I =x
+2+
ec
x
Q-3)1 – sin
1 – cos
x xe dx
x
∫∫∫∫
Ans. Let I =x 1– sin
1 – cos
xe dx
x
∫
I = 2
x
1 – 2sin .cos2 2
2sin2
x x
e dxx
∫
I =2x1
cos – 2cot2 2 2
x xe ec dx
∫
I =2x1
–2cot + cos2 2 2
x xe ec dx
∫
put f(x) = –2 cot 2
x
than f ′′′′(x) = cosec2
2
x
so,
I = ( ) ( )'x1+
2e f x f x ∫
I = ( )x1+
2e f x c
I =x1–2cot +
2 2
xe c
⇒⇒⇒⇒ I =x– cot +
2
xe c
Q-4)
(((( ))))
2
32 2
– +1
+1
x x xe dx
x∫∫∫∫
Ans. Let I =
( )
2
32 2
x – +1
+1
x xe dx
x∫
I =( )
( )
2
32 2
x+1 –
+1
x xe dx
x
∫
I =( )
( ) ( )
2
3 32 22 2
x+1
–
+1 +1
x xe dx
x x
∫
Mahesh Tutorials Science 57
Indefinite Integration
I =
( ) ( )1 3
2 22 2
x 1–
+1 +1
xe dx
x x
∫
f(x) =
( )( )
1–
2 21
2 2
1= +1
+1
x
x
Then f ′′′′(x) = ( ) ( )3
–2 22
1– +1 . +12
dx x
dx
=
( )3
2 2
2–
2 +1
x
x
=
( )3
2 2
–
+1
x
x
So,
I = ( ) ( )'x +e f x f x dx ∫I = ex f(x) + c
=
( )1
2 2
x 1+
+1
e c
x
I = 2
x
++1
ec
x
Q-5) (((( )))) (((( ))))sin log + cos logx x dx ∫∫∫∫
Ans. Let I = ( ) ( )sin log + cos logx x dx ∫Put x = t ⇒⇒⇒⇒ x = et ⇒⇒⇒⇒ dx = et dt
I = [ ] tsin + cost t e dt∫Put f (t) = sin t ⇒⇒⇒⇒ f ′′′′(t) = cos t
I = ( ) ( )'t +e f t f t dt ∫= et f (t) + c
= et sin t + c
I = x sin (log x) + c
Q-6)(((( ))))2
–1tan
2
1+ +
1+
xx x
e dxx∫∫∫∫
Ans. Let I = ( )2
-1tan
2
x1+ +
1+
x xe dx
x∫Put tan–1 x = t ⇒⇒⇒⇒ x = tan t
∴∴∴∴ 2
1=
1+
dxdt
x
I = ( )2t 1+ tan + tane t t dt∫I = ( )2t sec + tane t t dt∫I = ( )2t tan +sece t t dt∫Let f(t) = tan t ⇒⇒⇒⇒ f ′′′′(t) = sec2 t
I = ( ) ( ){ }'t +e f t f t dt∫So,
I = et f (t) + c
I =-1tan x +e x c [ ]tan =t x∵
I =-1tan x +xe x c
Q-7) (((( ))))2
log
1+ log
xdx
x∫∫∫∫
Ans. Let I = ( )2
log
1+ log
xdx
x∫Put log x = t
Then x = et ⇒⇒⇒⇒ dx = etdt
I = ( )2
t
1+
te dt
t∫
I =( )
( )2
t1+ –1
1+
te dt
t
∫
I =( )
2
t 1 1–
1+ 1+e dt
t t
∫
If ( )1
=1+
f tt
Indefinite Integration
58 Mahesh Tutorials Science
Then ( )( )
2'
–1=1+
f tt
I = ( ) ( )'t +e f t f t dt ∫I = et f(t) + c
I =t 1. +1+
e ct
I = +1+ log
xc
x
Q-8)(((( ))))
2
2
1
1
x xe dx
x
++++
++++ ∫∫∫∫
Ans. I =( )
2
2
x 1+
1+
xe dx
x
∫
=( )
2
2
x +1+ 2 – 2
+1
x x xe dx
x
∫
=( )
( )
2
2
+1 – 2
+1
xx x
e dxx
∫
=( )
2
x x– 2+1
xe dx e dx
x∫ ∫
=( )
( )x x
+1 –1– 2
+1
xe e dx
x∫
I =( )
2
x x1 –1– 2 +
+1 +1e e dx
x x
∫
Let ( ) ( )( )
1 –1= =
+1 +1f x f x
x x′
2
I = ( ) ( )( )– 2 +e e f x f x dx
′ ∫x x
= ( )x x– 2 +e e f x c
=
xx 2– +
+1
ee c
x
=x–1+
+1
xe c
x
I =x –1
++1
xe c
x
Q-9)(((( ))))
2
2
log –1
1 log
xdx
x
++++ ∫∫∫∫
Ans. I =( )
2
2
log –1
1+ log
xdx
x
∫
Put log x = t ∴∴∴∴ x = et
∴∴∴∴ dx = et dt
∴∴∴∴ I =
2
2
t–1
1+
te dt
t
∫
=( )
2
22
t – 2 +1
1+
t te dt
t
∫
=
( )
( ) ( )
2
2 22 2
t1+ 2
–1+ 1+
t te dt
t t
∫
=( )
2 22
t 1 2–
1+ 1+
te dt
t t
∫
Now, 2
1
1+
d
dt t
= ( )–1
21+d
tdt
= ( ) ( ) ( )–1–1
2 2–1 1+ 1+d
t tdx
= ( ) ( ) ( )–2
2–1 1+ 0 + 2t t
=( )
22
–2
1+
t
t
∴∴∴∴ I = 2 2
t 1 1+
1+ 1+
de dt
dtt t
∫
= 2
t 1+
1+e c
t
=( )
2
1+
1+ logx c
x
I = ( )2+
1+ log
xc
x
Mahesh Tutorials Science 59
Indefinite Integration
Q-6)
2
2 2
( – 1)
( +1)
x xe dx
x∫∫∫∫
Ans. =
2
4 2
x – 2 +1
+ 2 +1
x xe dx
x x
∫
=
2
2 2 2 2
x +1 2–
( +1) ( +1)
x xe dx
x x
∫
= 2 2 2
x 1 2–
1 ( +1)
xe dx
x x
∫ +
Let, f(x) = 2
1
+1x
f′ (x) =( )
2
–1 2
+1
x
x
( ) ( ) ( )'x x∴∴∴∴ + = +e f x f x dx e f x c
∫= 2
x
+1
e
x + c
GROUP (L) – HOME WORK PROBLEMS
Q-1) (sin + cos ) xe x x dx∫∫∫∫Ans. Let, f(x) = sin x
∴∴∴∴ f′ (x) = cos x
= exsinx + c
( ) ( ) ( )'x x∴∴∴∴ + = +e f x f x dx e f x c
∫
Q-2) 2
cos + sin
cos
x x xe dx
x∫∫∫∫Ans. I = ( )x sec + tan sece x x x∫
Let, f(x) = sec x
∴∴∴∴ f′ (x) = tan x.,
I = = ex sec x + c
( ) ( ) ( )'x x∴∴∴∴ + = +e f x f x dx e f x c
∫
Q-3) 2 + sin2
1+ cos2
x xe dx
x
∫∫∫∫
Ans. I = 2
x 2+2sin cos
2cos
x xe dx
x
∫
= ( )2x sec + tane x x dx∫Let, f(x) = tan x
f′ (x) = sec2 x
I = ex tanx + c
( ) ( ) ( )+ = +e f x f x dx e f x c
∫ 'x x∵∵∵∵
Q-4) 2 – sin2
1 – cos2
x xe dx
x
∫∫∫∫
Ans. I = 2
x 2 – 2sin cos2
2sin
x xe dx
x
∫
= ( )2x cos – cote ec x x dx∫Let, f(x) = – cot x
f′ (x) = cosec2 x
I = – ex cot x + c
( ) ( ) ( )+ = +e f x f x dx e f x c
∫ 'x x∵∵∵∵
Q-5) 1+ sin2
41+ cos2
x xe dx
x
∫∫∫∫
Ans. put 2x = t
∴∴∴∴ 2 dx = dt
Let, I = 2x 1+ sin2
1+cos2
xe dx
x
∫
=t ××××1+sin
1+cos 2
t dte
t
∫
=2 2
t
2sin cot1 1 2 2 +2
2cos 2cos2 2
t t
e dtt t
∫
=2t1 1
sec + tan2 2 2 2
t te dt
∫
Let, f(x) = tan 2
t
f′ (x) = sec2 1.
2 2
t
( ) ( ) ( )'x x∴∴∴∴ + = . +e f x f x dx e f x c
∫=
t1tan +
2 2
te c
=2x1tan +
2e x c
Indefinite Integration
60 Mahesh Tutorials Science
Q-7) 2( +1)
xxedx
x∫∫∫∫
Ans. = 2 2
x +1 1–
( +1) ( +1)
xe dx
x x
∫
= 2
x 1 1–
+1 ( +1)e dx
x x
∫
Let, f(x) = 1
1+ x
f′ (x) = ( )2
–1
1+ x
∴∴∴∴ I =x
+1
e
x + c
Q-8) 1
+ logxe x dxx
∫∫∫∫
Ans. put f(x) = log x
∴∴∴∴ f’ (x) = 1
x
∴∴∴∴ I = ( ) ( )'x +e f x f x dx ∫= ex f(x) + c
I = ex log x + c
Q-9) (tan – log cos )xe x x dx∫∫∫∫Ans. f(x) = –log(cos x)
∴∴∴∴ f′(x) =–1
cos x ·(–sinx) = tanx
( ) ( ) ( )'x x+ = +e f x f x dx e f x c ∫ = – ex (log cosx) + C
Q-10) (((( ))))cot + log sinxe x x dx∫∫∫∫
Ans. Let I = ( )x cot + log sine x x dx∫Put f(x) = log sinx. Then
f ′(x) = d
dx(log sin x) =
1
sinx. cos x = cot x
∴∴∴∴ I = [ ]'x ( )+ ( )e f x f x dx∫= ex.f(x) + c
= ex.log sin x + c
Q-11) (((( ))))sin sin cos cosxe x x x dx++++∫∫∫∫
Ans. I = ( )sin x cos sin +1e x dx∫put sinx = t ∴∴∴∴ cos x dx = dt
∴∴∴∴ I = ( )t +1e t dt∫f ′ (t) = t f ′ (t) = 1
I = et . t + c
I = esinx . sin x + c
Q-12)2
–1sin
2
1 –
1 –
x x xe dx
x
++++ ∫∫∫∫
Ans. Put sin–1 x = y ∴∴∴∴ x = sin y
∴∴∴∴ 2
1
1 – xdx = dy
I =2y sin + 1 – sine y y dy
∫= [ ]y sin + cose y y dy∫
f (y)= sin y f ′ (y) = cos y
I = [ ]y sin + cose y y dy∫= ey sin y + c
I = esin–1x sin (sin–1 x) + c
I = esin–1x (x) + c
GROUP (M) – CLASS WORK PROBLEMS
Q-1) 2
1
– 3 + 2dx
x x∫∫∫∫
Ans. Let I = 2
1
– 3 +2dx
x x∫
= ( ) ( )
1
– 2 –1∫ dxx x ...(i)
Let ( ) ( )
1
– 2 –1∫ dxx x
= +–1 – 2
A B
x x
Mahesh Tutorials Science 61
Indefinite Integration
Q-4) 2
sec2
cos + 4cos + 3
xdx
x x∫∫∫∫Ans. Let I = 2
sec2
cos + 4cos +3
xdx
x x∫= 2
2sin .cos
cos + 4cos +3
x xdx
x x∫Put cos x = t
∴∴∴∴ sinx dx = –dt
∴∴∴∴ I = 2
–2
+ 4 +3
t dt
t t∫= ( ) ( )
–2+ 3 +1
t dt
t t∫
Let ( ) ( )+ 3 +1
t
t t
= ++ 3 +1
A B
t t
∴∴∴∴ A == –3t+1
t
t
=–3
–3 +1=
3
2
B == –1t+ 3
t
t
=–1
–1+ 3=
1–2
∴∴∴∴ I =
–13
22–2 ++3 +1
dtt t
∫
=3 1
–2 log + 3 – log +1 +2 2
t t c
= log|cos x + 1| – 3 log|cos x + 3| + c
Q-2) 2
3 – 4
– 3 + 2
xdx
x x∫∫∫∫
Ans. Let I = 2
3 – 4
– 3 +2
xdx
x x∫
I = ( ) ( )
3 – 4
–1 – 2
xdx
x x∫
Let ( ) ( )
3 – 4
–1 – 2
x
x x = +–1 – 2
A B
x x
∴∴∴∴ 3x – 4 = A (x – 2) + B(x – 1)
Put x = 1 ⇒⇒⇒⇒ – 1 = A (–1) ⇒⇒⇒⇒ A = 1
Put x = 2 ⇒⇒⇒⇒ – 2 = B (1) ⇒⇒⇒⇒ B = 2
I = ( ) ( )
3 – 4
–1 – 2
xdx
x x∫
I = 1 2
+–1 – 2
dxx x
∫
I = log|x – 1| + 2 log |x – 2| + c
Q-3) (((( )))) (((( ))))
2sec
2 + tan 3 + tan
xdx
x x∫∫∫∫
Ans. Let I = ( ) ( )
2sec
2+ tan 3+ tan
xdx
x x∫Put tan x = t
∴∴∴∴ sec2x dx = dt
∴∴∴∴ I = ( ) ( )2+ 3 +
dt
t t∫ ... (i)
Let ( ) ( )2+ 3+
dt
t t
= +2+ 3 +
A B
t t... (ii)
∴∴∴∴ A ==1x
1= –1
– 2x
B ==2x
1=1
–1x
∴∴∴∴ I =–1 1
+–1 – 2
∫ dxx x
= –log|x – 1| + log |x – 2| + c
∴∴∴∴ I =– 2
log +–1
xc
x
∴∴∴∴ A == –2t
1=1
3+ t
B == –3t
1= –1
2+ t
∴∴∴∴ A = 1 and B = – 1 ...(iii)
∴∴∴∴ From (i), (ii), and (iii)
∴∴∴∴ I =( )–11
+2+ 3+
dtt t
∫
= log|2 + t| – log |3 + t| + c
=2+
log +3+
tc
t
=2+ tan
log +3+ tan
xc
x
Indefinite Integration
62 Mahesh Tutorials Science
Q-5)
2
2
–
– 2 – 3
x xdx
x x∫∫∫∫Ans. Let I =
2
2
–
– 2 – 3
x xdx
x x∫= ( )
2
2
––1
– 2 – 3
x xdx
x x∫= ( )
2
2
– 2 + – 3 +3–1
– 2 – 3
x x xdx
x x∫= ( )
2
2 2
– 2 – 3 +3–1 –
– 2 – 3 – 2 +3
x x xdx dx
x x x x∫ ∫= ( )
2
1 2 +6–1 –
2 – 2 +3
xdx dx
x x∫ ∫= ( )
2
1 2 – 2+8–1 –
2 – 2 – 3
xdx dx
x x∫ ∫= ( )
2
2
1 2 – 2–1 –
2 – 2 – 3
1– 8
– 2 – 3
xx dx
x x
dxx x
∫∫
= 2
2
1– – log – 2 – 3
2
1– 8
– 2 +1 –1 – 3
x x x
dxx x∫
=( ) ( )
2
2 2
1– – log – 2 – 3 – 8
2 –1 – 2
dxx x x
x∫
=2
××××××××
1– – log – 2 – 3
2
1 –1 – 2– 8 log +
2 2 –1+ 2
x x x
xc
x
∴∴∴∴ I =21 – 3
– – log – 2 – 3 – 2log +2 –1
xx x x c
x
Q-6)
2
2 2
+10 + 21
xdx
x x∫∫∫∫Ans. Let I =
2
4 2
+10 + 21
xdx
x x∫ ... (i)
Put x2 = t
∴∴∴∴
2
4 2+10 +21
x
x x= 2 +10 + 21
t
t t
= ( ) ( )+ 7 + 3
t
t t
= ++ 7 + 3
A B
t t
∴∴∴∴ A == –7t+3
t
t
=–7
–4
=7
4
∴∴∴∴ B == –3t+ 7
t
t
=–3
–3 + 7
=–3
4
∴∴∴∴
2
4 2+10 +21
x
x x=
7 –3
4 4++ 7 +3t t
=2 2
7 –3
4 4++ 7 +3x x
... (ii)
∴∴∴∴ From (i) and (ii),
∴∴∴∴ I = 2 2
7 3
4 4–+ 7 +3
dxx x
∫
= 2 2
7 1 3–
4 4+7 +3
dxdx
x x∫ ∫=
( ) ( )2 2
2 2
7 3–
4 4+ 7 + 3
dx dx
x x∫ ∫
= –1
–1
××××
××××
7 1tan
4 7 7
3 1– tan +4 3 3
x
xc
∴∴∴∴ I =–1 –17 3
tan – tan +4 47 3
x xc
Q-7) (((( )))) (((( ))))
2
2 2
2 –1
+ 4 – 5
xdx
x x∫∫∫∫Ans. Let I =
( ) ( )
2
2 2
2 –1
+ 4 – 5
xdx
x x∫ ... (i)
Put x2 = t
∴∴∴∴( ) ( )
2
2 2
2 –1
+ 4 – 5
x
x x= ( ) ( )
2 –1
+ 4 – 5
t
t t
= ++ 4 – 5
A B
t t
∴∴∴∴ A == –4t
2 –1=1
– 5
t
t
Mahesh Tutorials Science 63
Indefinite Integration
Q-8)(((( )))) (((( ))))
2– 1 +1
dx
x x∫∫∫∫Ans. Let I =
( ) ( )2
–1 +1
dx
x x∫ ... (i)
Let
( ) ( )2
1
–1 +1x x= ( ) ( ) ( )2
+ +–1 +1–1
A B C
x xx
=( ) ( ) ( ) ( )
( ) ( )
2
2
–1 +1 + +1 + –1
–1 +1
A x x B x C x
x x
⇒⇒⇒⇒ A(x2 – 1) + B (x + 1) + C(x – 1)2 = 1
Put x = 1
∴∴∴∴ 2B = 1
∴∴∴∴ B = 1
2
Put x = –1
∴∴∴∴ 4C = 1
∴∴∴∴ C = 1
4
Put x = 0
∴∴∴∴ –A + 1
2 + 1
4 = 1
∴∴∴∴ A = 1–4
∴∴∴∴( ) ( )
2
1
–1 +1x x=
( ) ( ) ( )2
11 1–
24 4+ +–1 +1–1x xx
.... (ii)
Q-9)(((( )))) (((( ))))
2
2
5
+1 + 4
xdx
x x∫∫∫∫
Ans. Let I = ( ) ( )
2
2
5
+1 + 4
xdx
x x∫ ... (i)
Let
( ) ( )
2
2
5
+1 + 4
x
x x =( ) ( )2
++
–1 + 4
A Bx c
x x
... (ii)
( ) ( )
2
2
5
+1 + 4
x
x x = ( ) ( ) ( )
( ) ( )
4 + + +1
+1 + 4
A x Bx C x
x x
+2
2
∴∴∴∴ A(x2 – 4) + (Bx + C)(x – 1) = 5x2
Put x = –1,
∴∴∴∴ 5A + (C – B)(O) = 5
∴∴∴∴ A = 1
Put x = 0,
∴∴∴∴ 4A + C = 0
∴∴∴∴ C = –4A
= –4(1) = –4
Put x = 1
∴∴∴∴ 5A + 2(B + C) = 5
∴∴∴∴ 5(1) + 2(B – 4) = 5
∴∴∴∴ 2(B – 4) = 0
∴∴∴∴ B = 4
∴∴∴∴( ) ( )
2
2
5
+1 + 4
x
x x= 2
1 4 + 4+
–1 + 4
x
x x
.... (iii)
∴∴∴∴ From (i) and (iii)
∴∴∴∴ I = 2
1 4 – 4+
+1 – 4
xdx
x x
∫
=
( ) ( )
1 2+2
+1 – 4
1– 4
+ 2
xdx dx
x x
dxx
∫ ∫∫
2
22
∴∴∴∴ B ==5t
2 –1=1
+ 4
t
t
∴∴∴∴( ) ( )
2
2 2
2 –1
+ 4 – 5
x
x x=
1 1+
+ 4 – 5t t
= 2 2
1 1+
+ 4 – 5x x... (ii)
∴∴∴∴ From (i) and (ii),
∴∴∴∴ I = 2 2
1 1+
+ 4 +5dx
x x
∫
= ( ) ( )2 22 2
++ 2 – 5
dx dxdx
x x∫ ∫
=–11 1 – 5
tan + log +2 2 2 5 + 5
x xc
x
∴∴∴∴ From (i) and (ii)
∴∴∴∴ I =( ) ( ) ( )2
1–1 1
24 4+ +–1 –1–1
dxx xx
∫
= ( )
1 1 1– log –1 – + log +1 +4 2 –1 4
x x cx
= ( )
1 +1 1log – +
4 –1 2 –1
xc
x x
Indefinite Integration
64 Mahesh Tutorials Science
Q-10) (((( ))))
1
cos 2 + sindx
x x∫∫∫∫Ans. Let I = ( )cos 2 + sin
dx
x x∫ ... (i)
Put sin x = t
∴∴∴∴ cos x dx = dt
∴∴∴∴ dx = cos
dx
x.
∴∴∴∴ I =( )
cos
cos 2+
dt
x
x t
∫
= ( )2cos 2+
dt
x t∫
= ( ) ( )21 – sin 2+
dt
x t∫
= ( ) ( )21 – 2+
dt
t t∫
= ( ) ( ) ( )1+ 1 – 2+
dt
t t t∫ ... (ii)
Let ( ) ( ) ( )
1
1+ 1 – 2+t t t
= + +1+ 1 – 2+
A B C
t t t
∴∴∴∴ A = ( ) ( )= –1t
1
1 – 2+t t
=1
2
B = ( ) ( )=1t
1
1+ 2+t t
=1
6
C = ( ) ( )= –2t
1
1+ 1 –t t
=1–3
∴∴∴∴ ( ) ( ) ( )
1
1+ 1 – 2+t t t=
1 1 –1
2 6 3+ +1+ 1 – 2+t t t
... (iii)
∴∴∴∴ I = log|x – 1| + 2 log|x2 + 4|
– 4 × × × × –11
tan +2 2
xc
= log|x – 1| + 2 log|x2 + 4|
– –12tan + .2
xc
∴∴∴∴ From (ii) and (iii)
∴∴∴∴ I =
1 1 –1
2 6 3+ +1+ 1 – 2+
dtt t t
∫
=1 1 1log 1+ – log 1+ – log 2+ +
2 6 3t t t c
∴∴∴∴ I =1 1log 1+ sin – log 1 – sin
2 6
1– log 2+ sin +3
x x
x c
GROUP (M) – HOME WORK PROBLEMS
Q-1) 2
1
– 5 + 6dx
x x∫∫∫∫Ans. Let I = 2
1
– 5 +6dx
x x∫= ( ) ( )
1
– 3 – 2dx
x x∫ ...(i)
Let ( ) ( )
1
– 3 – 2x x∫
= ( ) ( )+
– 3 – 2
A B
x x
∴∴∴∴ A ==3x
1=1
– 2x
∴∴∴∴ B ==2x
1= –1
– 3x
∴∴∴∴( ) ( )
1
– 3 – 2x x∫ =( ) ( )
1 1–
– 3 – 2x x
....(ii)
∴∴∴∴ From (i) and (ii)
∴∴∴∴ I =( )
( )
–11+
– 3 – 2dx
x x
∫
=1 1
–– 3 – 2
dx dxx x∫ ∫
= –log|x – 3| + log |x – 2| + c
=– 3
log +– 2
xc
x
Mahesh Tutorials Science 65
Indefinite Integration
Q-2) 2
3 – 2
– 3 + 2
xdx
x x∫∫∫∫Ans. Let = ( ) ( )
3 – 2
– 2 –1
xdx
x x∫3x – 2 = A(x – 1) + B(x – 2)
x = 2, x = 1
A = 4; –B = 1
B = – 1
=4 –1
+– 2 –1dx dx
x x∫ ∫= 4 log |x – 4| – log |x – 1| + c
Q-3) (((( )))) (((( ))))
log
2 + log 3 + logdx
x x x∫∫∫∫Ans. put log x = t
1
xdx = dt
+2+ 3 +
A B
t t = ( ) ( )2+ 3+
t
t t
A (3 + t) + B(2 + t) = t
put t = –2, t = –3
A = – 2 B = 3
I =–2 3
+2+ 3+
dt dtt t∫ ∫
I = –2log 1+ +3log 3+t t
I = 3 log |3 + log x| – 2log |2 + log x| + c
=( )
( )
3
2
3 + loglog +
2+ log
xc
x
Q-4) 2
sin2
sin – 5sin + 6
xdx
x x∫∫∫∫Ans. Let I = 2
sin2
sin – 5sin +6
xdx
x x∫= 2
2sin .cos
sin – 5sin +6
x x dx
x x∫put sinx = t
∴∴∴∴ cosx dx = dt
∴∴∴∴ I = 2
2
– 5 +6
t dt
t t∫=
( ) ( )
2
– 3 – 2
tdt
t t∫ ...(i)
Let ( ) ( )
2
– 3 – 2
tdt
t t∫
Q-5)
2
2
+1
– 4 + 3
xdx
x x∫∫∫∫Ans. Let =
2
2
+1
– 4 + 3
xdx
x x∫
2 2
2
+ +
1
– 4 – 3 +1
– 4 –3
4 – 2
x x x
x x
x
∴∴∴∴ I = 2
4 – 21+
– 4 +3
xdx
x x
∫
∴∴∴∴ I = 2
4 – 2+
– 4 +3
xdx dx
x x∫ ∫ ....(i)
4x – 2 = Ad
dx(x2 – 4x + 3) + B
= A(2x – 4) + B .....(ii)
= 2Ax – 4A + B
∴∴∴∴ 2A = 4
∴∴∴∴ A = 2
∴∴∴∴ –4A + B = – 2 ⇒⇒⇒⇒ B = 6
∴∴∴∴ 4x – 2 = 2(2x – 4) + 6 .....(iii)
∴∴∴∴ From (i) and (iii)
∴∴∴∴ I =( )2
2 2 – 4 +6+
– 4 + 3
xdx dx
x x
∫ ∫= 2 2
2 – 4+2 +6
– 4 +3 – 4 +3
x dxx dx
x x x x∫ ∫=
2
2+2log – 4 +3 +6
– 4 + 4 –1
dxx x x
x x∫= x + 2 log |x2 – 4x + 3|
××××1 – 2 –1
+6 log +2 – 2+1
xc
x
= ( ) ( )+
– 3 – 2
A B
t t
∴∴∴∴ A ==3t
2= 6
– 2
t
t
∴∴∴∴ B ==2t
2= – 4
– 3
t
t
∴∴∴∴( ) ( )
2
– 3 – 2
t
t t=
( )
( )
( )
–46+
– 3 – 2t t
∴∴∴∴ I =6 4
–– 3 – 2
dtt t
∫
= 6log|t – 3| – 4 log |t – 2| + c
= 6log|t – 3| – 4 log |sinx – 2| + c
Indefinite Integration
66 Mahesh Tutorials Science
∴∴∴∴ I = x + 2 log |x2 – 4x + 3|
+ 3 log – 3
–1
x
x+ c
∴∴∴∴ I = x + 2 log |(x – 3)(x –1)| + 3log – 3
–1
x
x+ c
∴∴∴∴ I = x + log ( ) ( )( )
( )
3
2 2
3××××
– 3– 3 –1 +
–1
xx x c
x
∴∴∴∴ I = x + log ( )
( )
5– 3
+–1
xc
x
∴∴∴∴ I = x + 5 log |x – 3| – log|x – 1| + c
Q-6) (((( )))) (((( ))))
2
2 2
+ 4 + 9
xdx
x x∫∫∫∫Ans. Let I =
( ) ( )
2
2 2
+ 4 + 9
xdx
x x∫ ... (i)
Let x2 = t
∴∴∴∴( ) ( )
2
2 2+ 4 + 9
x
x x=
( ) ( )+ 4 + 9
t
t t
= ++ 4 +9
A B
t t
∴∴∴∴ A == –4t+ 9
t
t
=4–5
∴∴∴∴ B == –9t+ 4
t
t
=9
5
∴∴∴∴( ) ( )
2
2 2+ 4 + 9
x
x x=
( ) ( )+ 4 + 9
t
t t
= ++ 4 + 9
A B
t t
∴∴∴∴ A ==4t
4= –
+ 9 5
t
t
∴∴∴∴ B == –9t
9=
+ 4 5
t
t
∴∴∴∴ ( ) ( )
2
2 2+ 4 + 9
x
x x =
–4 9
5 5++ 4 + 9t t
=2 2
–4 9
5 5++ 4 + 9x x
...(iii)
∴∴∴∴ From (i) and (iii)
∴∴∴∴ I = 2 2
–4 9
5 5++ 4 +9
dxx x
∫
= ( )2 22
–4 9+
5 5+ 4 + 3
dx dx
x x∫ ∫
=–1 –1× ×× ×× ×× ×
–4 1 9 1tan + tan +
5 2 2 5 3 3
x xc
∴∴∴∴ I =–1 –1–2 3
tan + tan +5 2 5 3
x xc
Q-7) (((( )))) (((( ))))
2
2 2
2 – 3
+1 – 4
xdx
x x∫∫∫∫
Ans. Let I = ( ) ( )
2
2 2
2 – 3
+ 4 – 9
xdx
x x∫ ... (i)
Let x2 = t
∴∴∴∴ ( ) ( )
2
2 2
2 – 3
+1 – 4
x
x x = ( ) ( )
2 – 3
+1 – 4
t
t t
= ++1 – 4
A B
t t
∴∴∴∴ A == –1t
2 – 3
– 4
t
t
=–2 – 3
= +1–1 – 4
∴∴∴∴ B ==4t
2 – 3
+1
t
t
=( )2 4 – 3 5
= =14+1 5
∴∴∴∴ ( ) ( )
2
2 2
2 – 3
+1 – 4
x
x x =1 1+
+1 – 4t t
= 2 2
1 1+
+1 + 4x x
....(ii)
∴∴∴∴ From (i) and (ii)
∴∴∴∴ I = 2 2
1 1+
+1 – 4dx
x x
= 2 2 2
1+
+1 – 2
dxdx
x x∫ ∫= ( )
( )–1 1 – 2
tan + log +2 2 + 2
xx c
x
∴∴∴∴ I = ( )–1 1 – 2tan + log +
4 +2
xx c
x
Mahesh Tutorials Science 67
Indefinite Integration
Q-8)(((( )))) (((( ))))2 2+1 – 4
x dx
x x∫∫∫∫
Ans. Let I = ( ) ( )2 2+1 – 4
x dx
x x∫Put x2 = t
∴∴∴∴ 2x dx = dt
∴∴∴∴ x dx = 2
dt
∴∴∴∴ I =( ) ( )
2
+1 – 4
dt
t t
∫
= ( ) ( )
1
2 +1 – 4
dt
t t∫ ... (i)
Let ( ) ( )
1
+1 – 4t t∫= +
+1 – 4
A B
t t
∴∴∴∴ A == –1t
1
– 4t
=1–5
∴∴∴∴ B ==4t
1
+1t
=1
5
∴∴∴∴ ( ) ( )1
+1 – 4t t
=
–1 –1
5 5++1 – 4t t
....(ii)
∴∴∴∴ From (i) and (ii)
∴∴∴∴ I =
–1 11 5 5+2 +1 – 4
dtt t
∫
=1 1 1– log +1 + logg – 4 +
2 5 5t t c
=2 21
log – 4 – log +1 +10
x x c
Q-9) (((( )))) (((( ))))2– 1 +1
xdx
x x∫∫∫∫
Ans. Let I = ( ) ( )2–1 +1
xdx
x x∫ .... (i)
Let ( ) ( )2–1 +1
x
x x
=( ) ( ) ( )
( ) ( )
2
2
+1 + + –1
–1 +1
A x Bx c x
x x
∴∴∴∴ x = A(x2 + 1) + (Bx + c)(x – 1)
Put x = 1
∴∴∴∴ 2A = 1
∴∴∴∴ A = 1
2
Put x = 0
∴∴∴∴ A – C = 0
∴∴∴∴ A = C ⇒⇒⇒⇒ C = 1
2
Put x = –1
∴∴∴∴ 2A + (C –B) (–2) = –1
∴∴∴∴ 1 – 21–
2B
= –1
∴∴∴∴ 1 – 1 – 2B = –1 ⇒⇒⇒⇒ 2B = 1
∴∴∴∴ B = 1
2
∴∴∴∴ ( ) ( )2–1 +1
x
x x
=2
1 11 +2 22 +
–1 +1
x
x x
....(ii)
∴∴∴∴ From (i) and (ii)
I = 2
1 11 +2 22 +
–1 +1
x
dxx x
∫
= 2 2 2
1 1 2 1+ +
2 –1 22 +1 +1
dx x dxdx
x x x x∫ ∫ ∫=
( )
2
–1
1 1log –1 + log –1
2 4
1+ tan +2
x x
x c
Indefinite Integration
68 Mahesh Tutorials Science
Q-10)sin sin2
dx
x x++++∫∫∫∫Ans. Let I =
sin + sin2
dx
x x∫I =
sin +2sin cos
dx
x x x∫I = ( )sin 1+ 2cos
dx
x x∫
I = ( )2
sin
sin 1+ 2cos
xdx
x x∫
I = ( ) ( )2
sin
1 – cos 1+2cos
x dx
x x∫Put t = cos x ⇒⇒⇒⇒ – sin x dx = dt
I = ( ) ( )2
–
1 – 1+ 2
dt
t t∫
I =( ) ( ) ( )
–1+ 1 – 1+ 2
dt
t t t∫Let + +
1+ 1 – 1+ 2
A B C
t t t
= ( ) ( ) ( )
1
1+ 1 – 1+ 2t t t
A(1 – t)(1 + 2t) + B(1 + 2t)(1 + t)
+ C(1 + t)(1 – t) = 1
Put t = –1⇒⇒⇒⇒ A(1 + 1)(1 – 2) = 1
⇒⇒⇒⇒ A = –1
2
Put t = –1⇒⇒⇒⇒ B(1 + 2)(1 + 1) = 1
⇒⇒⇒⇒ B = 1
6
Put t = –1
2 ⇒⇒⇒⇒
1 11 – 1+
2 2C
= 1
⇒⇒⇒⇒ 1 3
2 2C
= 1 ⇒⇒⇒⇒ C = +4
3
∴∴∴∴ ( ) ( ) ( )
1
1+ 1 – 1+ 2t t t
=
–1 1 +4
2 6 3+ +1+ 1 – 1+2t t t
∴∴∴∴ I =
–1 1 4
2 6 3– + +
1+ 1 – 1+2dt
t t t
∫
I =
–1 1 4
2 6 3– + +
1+ 1 – 1+2dt
t t t
∫
I =log 1 –1 1
+ log 1+ –2 6 –1
tt
××××4 1
– log 1+ 2 +3 2
t c
=1 1
+ log 1+ cos + log 1 – cos2 6
x x
2– log 1+ 2cos +3
x c
Q-11)(((( ))))
2
2 1
– 4
xdx
x
++++
∫∫∫∫
Ans. Let ( )
2
2 + 7
– 4
x
x = ( ) ( )
2+
– 4 – 4
A B
x x
=( )
( )2
– 4 +
– 4
A x B
x
2x + 7 = A (x – 4)2 + B = Ax + (–4A + B)
On comparing the co-efficient
A = 2 and –4A + B = 7
∴∴∴∴ B = 7 + 4A = 7 + 8 =15
∴∴∴∴( )
2
2 + 7
– 4
x
x = ( ) ( )
2
2 15+
– 4 – 4x x
I =( )
2
152 +15
– 4 – 4
dx
x x∫ ∫
= 2 log |x – 4| + 15 + ( )
( )
–2+1– 4
–2+1
x+ c
I = 2 log |x – 4| – ( )
15
– 4x+ c
Q-12)(((( ))))
2
3
1
–1
x xdx
x
+ ++ ++ ++ +
∫∫∫∫
Ans. Let ( )
2
3
+ +1
–1
x x
x =
( ) ( ) ( )2 3
+ +–1 –1 –1
A B C
x x x
x2 + x + 1 = A (x – 1)2 + B (x –1)+ C ...(i)
Comparing the co-efficients we get
1 = A ∴ ∴ ∴ ∴ A = 1
Put x = 1 in (i), we get
∴∴∴∴ 3 = C ⇒⇒⇒⇒ C = 3
Mahesh Tutorials Science 69
Indefinite Integration
Q-13)(((( )))) (((( ))))
2
3 – 2
1 3
xdx
x x+ ++ ++ ++ +∫∫∫∫Ans. I =
( ) ( )2
3 – 2
+1 + 3
xdx
x x∫
( ) ( )2
3 – 2
+1 +3
x
x x=
( ) ( ) ( )2+ +
+1 +3+1
A B C
x xx
( ) ( )2
3 – 2
+1 +3
x
x x
= ( ) ( ) ( )
( ) ( )
2
2
+3 +1 +1
+1 +3
x A x + B C x
x x
+
∴∴∴∴ 3x – 2 = (x + 3) [A (x + 1) + B] + C (x + 1)2
when x = – 1
3 (–1) – 2 = (–1 + 3) [A (0) + B] + C (–2)2
–11 = 4C ∴∴∴∴ C = –11
4
when x = 0, 3 (0) – 2 = 3 [A (0 + 1) + B]
+ C [0 +1]2
– 2 = 3A + 3B + C
–2 = 3A + 5 11
3 – –2 4
3A =15 11
–2+ +2 4
= 33
4
∴∴∴∴ A = 11
4
∴∴∴∴ I = ( )
2
11/4 5/2 11/4– –
+1 + 3+1dx
x xx
∫
= ( )
2
11 5 11– –
4 +1 2 4 + 3+1
dx dx dx
x xx∫ ∫ ∫
I = 11 5 1
log +1 – –4 2 +1
xx
11– log + 3 +4
x c
Q-14)(((( )))) (((( ))))2
8
2 4dx
x x+ ++ ++ ++ +∫∫∫∫
Ans. I = ( ) ( )2
8
+ 2 + 4x x = 2
++
+ 2 + 4
A Bx C
x x
8 = A (x2 + 4) + (Bx + c) (x + 2) ...(i)
8 = (A + B)x2 + (2B + c)x (4A + 2C) ...(ii)
Putting x = – 2 in equation (i) we get
8 = 8A ⇒ A = 1
Comparing co-efficient of x2 and constant
term on both sides, we get
A + B = 0 and 4A + 2C = 8
B = – 1 and 4 + 2C = 8 ⇒ C = 2
( ) ( )2
8
+ 2 + 4x x = 2
1 – + 2+
+ 2 + 4
x
x x
∴∴∴∴ I = 2
1 – + 2+
+2 + 4
xdx
x x
∫
= 2
1 1 2–
+ 2 2 + 4
xdx dx
x x∫ ∫ 2 2
1+2
+2dx
x∫I = log |x + 2|
2 –11 1
– log + 4 +2. tan +2 2 2
xx c
I = –1
2
+2log + tan +
2+ 4
x xc
x
Q-15)
(((( )))) (((( ))))
2
2 2
37
– 7 4
++++
++++∫∫∫∫x
dxx x
Ans. Let I = ( ) ( )
2
2 2
+ 37
– 7 + 4
xdx
x x∫
= ( ) ( )
2
2 22 2
+ 37+
– 7 + 4– 7 + 4
x A B
x xx x= ...(i)
x2 + 37 = A (x2 + 4) + A (x2 + 4) ... (ii)
Putting x2 = 7 in (ii) we get
∴∴∴∴ 7 + 37 = A (7 + 4)
44 = 11 A
A = 4
Putting x2 = – 4 in (ii) we get
–4 + 37 = B (–4 – 7)
∴∴∴∴ Put x = 0 in (i), we get
1 = A – B + C ; 1 = 1 – B + 3
∴∴∴∴ B = 3
∴∴∴∴( )
2
3
+ +1
–1
x x
x=
( ) ( ) ( )2 3
+ +–1 –1 –1
A B C
x x x
I = ( ) ( )–2 –31
+3. –1 +3 –1–1
x x dxx
∫
I = log |x – 1| – ( ) ( )
2
3 3–
–1 2 –1x x
+ c
= ( )
11 +1 5log + +
4 + 3 2 +1
xc
x x
Indefinite Integration
70 Mahesh Tutorials Science
33 = –11 B
B = –3
∴∴∴∴ From (i) we get
( ) ( )
2
2 22 2
+ 37 4 –3= +
– 7 + 4– 7 + 4
x
x xx x
I = 2 2
1 14 – 3 =
– 7 + 4dx dx
x x∫ ∫=
( )2 2 2
2
1 14 – 3 =
+ 2– 7dx dx
xx∫ ∫
= –12 – 7 3log – tan +
2 27 + 7
x xc
x
GROUP (N) – CLASS WORK PROBLEMS
Q-1)5sin xdx∫∫∫∫
Ans. I =5sin x dx∫
=4sin sinx x dx∫
= ( )4
21 – cos .sinx x dx∫cos x = t
∴∴∴∴ –sin x dx = dt
∴∴∴∴ sin x dx = dt
∴∴∴∴ I = ( ) ( )2
21 – –t dt∫= ( )
22 41 – 2 +t t dt∫
=3 52 1
– – + +3 5
t t t c
=3 52 1
cos – cos + cos +3 5
x x x c
Q-2)4cos x dx∫∫∫∫
Ans. I =4cos x dx∫
= ( )2
2cos x dx∫=
21+ cos2
2
xdx
∫
=21
1+2cos2 + cos 24
x x dx ∫=
1 1+cos41+2cos2 +
4 2
xx dx
∫
=1 1 sin4
+ sin2 + + +4 2 4
xx x x c
∫
Q-3)5tan x dx∫∫∫∫
Ans. I =5tan x dx∫
=3 2tan tanx x dx∫
= ( )3 2tan sec –1x x dx∫=
3 2 3tan sec – tanx x dx x dx∫ ∫In the first integral, put tan x = t
∴∴∴∴ sec2 x dx = dt.
∴∴∴∴ I =3 2– tan .tant dt x x dx∫ ∫
= ( )4 2
1
1– tan sec –1 +
4t x x dx c∫
= ( )4 2
1
1tan – tan sec – tan +
4x x x x dx c∫
= 4 21
tan – tan sec4
x x x dx∫1+ tan +x dx c∫
Again, in the first integral, put tan x = t,
∴∴∴∴ sec2 x dx = dt.
∴∴∴∴ I =4
1 2
1– + log sec + +
4t x t dt x c c∫
=4 2
1 2 3
1 1– + log sec + + +
4 2t x t x c c c
=4 21 1
tan – tan + log sec +4 2
x x x c
Q-4)6cot x dx∫∫∫∫
Ans. I =6cot x dx∫
I = ( )4 2cot cos –1x ec x dx∫
Mahesh Tutorials Science 71
Indefinite Integration
Q-3)4sin x dx∫∫∫∫
Ans. Let I =4sin x dx∫
= ( )2
2sin x dx∫=
21 – cos2
2
xdx
∫
=21
1 – 2cos2 + cos 24
x x dx ∫=
1 1+ cos 41 – 2cos2 +
4 2
xx dx
∫
=1 2sin2 1 sin4
– + + +4 2 2 8
x xx x c
=1 sin43 – 2sin2 + +
8 4
xx x c
Q-5)6sec x dx∫∫∫∫
Ans. I =6sec x dx∫
I = ( )2
4 2 2 2sec .sec sec secx x dx x x dx∫ ∫I = ( )
22 21+ tan .secx x dx∫
Put tan x = t
∴∴∴∴ sec2 x dx = dt
∴∴∴∴ I = ( )2
21+ t dt∫= ( )2 41+2 +t t dt∫=
3 52 1+ + +3 5
t t t c
=3 52 1
tan + tan + tan +3 5
x x x c
Q-6)4cosec x dx∫∫∫∫
Ans. I =2 2cos .cosec x ec x dx∫
I = ( )2 21+cot .cosx ec x dx∫Put cot x = t ⇒⇒⇒⇒ cos ec x dx = –dt
∴∴∴∴ I = ( ) ( )21+ –t dt∫=
3
– + +3
tt c
∴∴∴∴ I = – cot x – 1
3 cot3x + c
GROUP (N) – HOME WORK PROBLEMS
Q-1)3sin x dx∫∫∫∫
Ans. I =3sin x dx∫ =
2sin .sinx x dx∫= ( )21– cos .sinx x dx∫
Put cos x = t
∴∴∴∴ sin x dx = –dt
∴∴∴∴ I = ( ) ( )21– –t dt∫ =
3
– +3
tt c
=
3cos– cos +
3
xx c
Q-2)5cos x dx∫∫∫∫
Ans. I =5cos x dx∫ = ( )
221 – cos cosx x dx∫
Put sin x = t ⇒⇒⇒⇒ cos x dx = dt
∴∴∴∴ I = ( )2
21 – t dt∫= ( )2 41– 2 +t t dt∫
I =3 52 1
sin – sin + sin +3 5
x x x c
I = ( )
4 2
2 2
cot cos –
cot cos –1
x ec xdx
x ec x
∫
I = ( )4 2 2
2
cot – cot cos
+ cot
x x ec x dx
x dx
∫∫
Put cot x = t in first integral, cosec2x dx = – dt
I = ( ) ( )2 4 2– + cos –1t t dt ec x dx∫ ∫I =
3 5cot cot– – cot – +
3 5
x xx x c
Indefinite Integration
72 Mahesh Tutorials Science
Q-6)5cot x dx∫∫∫∫
Ans. I =5cot x dx∫ =
3 2cot cotx x dx∫I = ( )3 2cot cos –1x ec x dx∫I = ( )3 2 2cot cos – cot cotx ec x x x dx∫
Q-4)4tan x dx∫∫∫∫
Ans. I =4tan x dx∫ =
2 2tan .tanx x dx∫= ( )2 2sec –1 .tanx x dx∫= ( )2 2 2sec .tan – sec –1x x dx x dx∫ ∫
In first integral,
put tan x = t ⇒⇒⇒⇒ sec2 x dx = dt.
∴∴∴∴ I =2 2– sec +t dt x dx dx∫ ∫ ∫
=
3
– tan + +3
tx x c
=
3tan– tan + +
3
xx x c
Q-5)4cot x dx∫∫∫∫
Ans. I =4cot x dx∫ =
2 2cot .cotx x dx∫= ( )2 2cos –1 cotec x x dx∫=
( )
2 2
2
cot .cos
– cos –1
x ec x dx
ec x dx
∫∫
Put cot x = t in first integral,
∴∴∴∴ –cosec2x dx = – dt
⇒⇒⇒⇒ cosec2x dx = – dt
∴∴∴∴ I = ( )2– – – cot – +t dt x x c∫=
3
– + cot + +3
tx x c
=31
– cot + cot + +3
x x x c
Q-7)4sec x dx∫∫∫∫
Ans. Let I =4sec x dx∫ = ( )2 2sin .sinx x dx∫
= ( )2 21+ tan .secx x dx∫Put tan x = t
∴∴∴∴ sin2 x dx = dt
∴∴∴∴ I = ( )21+ t dt∫ =
3
+ +3
tt c
∴∴∴∴ I =31
tan + tan + 33
x x
Q-8)6cosec x dx∫∫∫∫
Ans. I =6cosec x dx∫
= ( )2
2 2cos cosec x ec x dx∫I = ( )
22 21+ cot cosx ec x dx∫
Put cot x = t ⇒⇒⇒⇒ cosec2x dx = – dt
I = ( )2
21+ t dt∫ = ( )2 4– 1+2 +t t dt∫I =
3 32 1– cot + cot + cot +
3 5x x x c
I = ( )
3 2
2
cot cos
– cot cos –1
x ec xdx
ec x
∫
I = ( )3 2cot – cot cos
+ cot
x x ec x dx
x dx
∫∫
Put cot x = t in first integral cos ec2x dx =–dt
I = ( )3 – + log sin +t t dt x c∫I =
2 4cot cot– + log sin +
2 4
x xx c
Mahesh Tutorials Science 73
Indefinite Integration
BASIC ASSIGNMENTS (BA)
BA – 1
Q-1)1
1+ sin∫∫∫∫ dxx
Ans. I =( ) ( )
1 – sin
1+ sin 1 – sin∫x
dxx x
= 2
1– sin
1 – sin∫xdx
x
= 2
1 – sin
cos∫xdx
x
= ( )2sec – tan sec∫ x x x dx
=2sec – tan sec∫ ∫x dx x x dx
I = tan x – sec x + c
Q-5)tan
sec + tan∫∫∫∫x
dxx x
Ans. I = tan
sec + tan
xdx
x x∫= ××××
tan sec – tan
sec + tan sec – tan∫x x x
dxx x x x
=
2
2 2
sec tan – tan
sec – tan∫x x x
dxx x
= ( )2sec tan – sec –1 ∫ x x x dx
=2sec tan – sec + 1∫ ∫ ∫x x dx x dx dx
= sec x – tan x + x + c
Q-4) 2 2
cos2
sin .cos∫∫∫∫x
dxx x
Ans. =
2 2
2 2
cos – sin
sin .cos∫x x
dxx x
= 2 2
1 1–
sin cos
∫ dx
x x
=2 2cosec – sec∫ ∫x dx x dx
= – cot x – tan x + c
Q-2)
2
2
– 3
+1∫∫∫∫x
dxx
Ans. I =
2
2
– 3
+1∫x
dxx
=
2
2
+1 – 4
+1∫x
dxx
=
2
2 2
+1 4–
+1 +1
∫x
dxx x
= 2
1– 4
+1∫ ∫dx dxx
I = x – 4 tan–1 x + c
Q-3) 0cos2∫∫∫∫ x dx
Ans. I =2
cos180
π
∫ x dx
=ππππ
ππππ
180 2sin +
2 180∫x
c
I =
0
ππππ
90 sin2x + c
BA – 2
Q-1)1
++++∫∫∫∫ dxx x
Ans. Let I =1
+dx
x x∫=
( )1
+1dx
x x∫
I = ( )2
2 +1dx
x x∫f (x) = ( )+1x
f ′′′′(x) =1
2 x
∴∴∴∴ I = 2 log +1 +x c
Indefinite Integration
74 Mahesh Tutorials Science
Q-2) (((( ))))1
2 + log xx x∫∫∫∫
Ans. I =1
2 + log∫ dxx x
=1
2 + log∫ dxx x x
I = ( )
1
2+ log∫ dxx x
Put log x = t ∴∴∴∴1
xdx = dt
I =1
2+∫ dtt
= log 2+ +t c
I = log 2+ log +x c
Q-3)1 – tan
1+ tan
xdx
x∫∫∫∫Ans. I =
1 – tan
1+ tan
xdx
x∫
=cos – sin
sin + cos∫x x
dxx x
f (x) =sin x + cos x f ′′′′ (x) = cos x – sin x
I =( )
( )
f xdx
f x
′
∫= log cos +sin +x x c
Q-4)(((( ))))
(((( ))))
sin –
sin +
x adx
x a∫∫∫∫Ans. I =
( )
( )
sin –
sin
x adx
x a+∫
=( )
( )
sin + – 2
sin +
x a adx
x a∫
=( ) ( )
( )
sin + cos2 – cos + sin2
sin +
x a a x a adx
x a∫
= ( )cos2 – cot + sin2adx x a adx∫ ∫∴∴∴∴ I = x cos 2a – sin 2a log |sin (x + a) | + c
Q-5)–3
1
+dx
x x∫∫∫∫Ans. I =
1
+dx
x x∫ –3
=+1
xdx
x∫3
4
Put x4 = t 4x3 dx = dt
I =1 1
4 1+dtt∫
=1log 1+ +
4t c
=1log 1+ +
4x c4
BA – 3
Q-1)2
1
7 – 6 –dx
x x∫∫∫∫Ans. I =
1
7 – 6 –dx
x x∫ 2
=( )( )2
1
2 – + 2 3 + 9dx
x x∫
=
( ) ( )2 2
1
2 – + 3
dx
x∫
=–1 + 3
sin +2
xc
Q-2)3
4 – 5sindx
x∫∫∫∫Ans. I =
3
4 – 5sindx
x∫
Put tan2
xt=
x = 2 tan t
dx =2
1+
dt
t2
∴∴∴∴ sin x =2
1+
t
t2
=2
2
3 22
2 1+4 – 5
1+
dt
t t
t
∫
Mahesh Tutorials Science 75
Indefinite Integration
= 2
32
4+ 4 –10dt
t t∫= 2
32
4 –10 + 4dt
t t∫=
2
6 1
104– +14
dt
t t∫
=2
3 1
52– +1
2
dt
t t
∫
=2
3 1
5 25 92– 2 + –
4 16 16
dt
t t
∫
=2 2
3 1
2 5 3– –4 4
dt
t
∫
=
5 3– –
3 1 4 4log +5 332 – +24 44
tc
t
I =
tan – 42log +
2tan –12
x
cx
Q-3) –
1
3 +2x xdx
e e∫∫∫∫Ans. I =
1
3 +2dx
e e∫ –x x
=3 +2
edx
e∫x
2x
Put ex = t ∴∴∴∴ ex dx = dt
I =2
1 1
23+3
dt
t∫
=2
1 1
23+3
dt
t∫
=1 1
3 2+
3t
∫ 2
2
= –11 1tan +
3 2 2
3 3
tc
I =–1 x1 3
tan +26e c
Q-4) 2
1
2 – 3cosdx
x∫∫∫∫Ans. I =
1
2 – 3cosdx
x∫ 2
Divide by cos2 x
I =sec
2sec – 3
xdx
x∫2
2
= ( )sec
2 1+ tan – 3
xdx
x∫2
2
=sec
2+2tan – 3
xdx
x∫2
2
=sec
2tan –1
xdx
x∫2
2
Put tan x = t
∴∴∴∴ sec2x dx = dt
I =1
2 –1dt
t∫ 2
=
1 1
12–2
dt
t∫ 2
=
1 1
2 1–
2
dt
t
∫2
=
1–
1 1 2log +
12 1 +2 –22
t
c
t
∫
I =1 2 tan –1
log +2 2 2 tan +1
xc
x
Q-5) 2
sin
9 – sin
xdx
x∫∫∫∫Ans. I =
sin
9 – sin
xdx
x∫ 2
=sin
9 –1+cos
xdx
x∫ 2
I =sin
8+cos
xdx
x∫ 2
Put cos x = t –sin x dx = dt
I =–1
8+dt
t∫ 2
Indefinite Integration
76 Mahesh Tutorials Science
=( )
1–1
+ 2+ 2
dt
t∫ 2
2
= –1 log + +8 +t t c
2
I = – log cos + 8+ cos +x x c2
BA – 4
Q-1)
2
4 2
+14
+ – 2
xdx
x x∫∫∫∫Ans. Let I =
2
4 2
+14
+ – 2
xdx
x x∫
= ( ) ( )
2
2 2
+14
+ 2 –1
xdx
x x∫ .... (i)
( ) ( )
2
2 2
+14
+ 2 –1
x
x x= ( ) ( )
+14
+ 2 –1
t
t t
= ++ 2 –1
A B
t t
Where t = x2 and
∴∴∴∴ A = =–2t
+14= –4
–1
t
t
∴∴∴∴ B = =1t
+14= 5
+ 2
t
t
∴∴∴∴ ( ) ( )
2
2 2
+14
+ 2 –1
x
x x =–4 5
++ 2 –1t t
= 2 2
–4 5+
+ 2 –1x x ... (ii)
∴∴∴∴ From (i) and (ii)
∴∴∴∴ I = 2 2
–14 5+
+2 –1dx
x x
∫
=
( ) ( )2 222
1– 4 +5
– 1+ 2
dxdx
xx∫ ∫
= –1– 4 5 –1tan + log +
2 +12 2
x xc
x
Q-2)2
5 +2
– 3 + 2
xdx
x x∫∫∫∫
Ans. Let I =
2
2
+2
– 3 +2
xdx
x x∫
= ( ) ( )
5 + 2
– 2 –1
xdx
x x∫ .... (i)
Let ( ) ( )
5 + 2
– 2 –1
x
x x
= +– 2 –1
A B
x x
Where
∴∴∴∴ A = =2x
5 + 2=12
–1
x
x
∴∴∴∴ B = 5 +2
= –7– 2
x
x
=1x
∴∴∴∴ ( ) ( )
5 + 2
– 2 –1
x
x x =12 7
–– 2 –1x x
∴∴∴∴ I =12 7
–– 2 –1
dxx x
∫
= 12 log|x – 2| – 7 log |x – 1| + c
Q-3)1+ cos
xdx
x∫∫∫∫
Ans. Let I =1+cos
xdx
x∫=
22cos2
xdx
x
∫
=21
.sec2 2
xx dx
∫
= ( )2
tan1 2sec – 1
12 2
2
xx
x dx dx
∫ ∫
=
tan1 2
. – 2 tan12 2
2
x
xx dx
∫
Mahesh Tutorials Science 77
Indefinite Integration
=
2log sec21
2 .tan – +12 2
2
x
xx c
=
2log sec21
2 .tan – +12 2
2
x
xx c
= .tan – 2log sec +2 2
x xx c
Q-4)(((( ))))
2+1
xxe
dx
x∫∫∫∫
Ans. I =( )
2
x
+1
xedx
x∫
=( )
2
x +1 –1
+1
xe dx
x
∫
=( ) ( )
2 2
x +1 1–
+1 +1
xe dx
x x
∫
I =( ) ( )
2
x 1 1–
+1 +1e dx
x x
∫
f(x) = ( )1
+1x
f ′′′′(x) = ( )
2
–1
+1x
I = 1
. ++1
e cx
x
I = ++1
ec
x
x
Q-5) 29 – 16x dx∫∫∫∫
Ans. Let I =29 –16x dx∫
I =2 16
3 –9
x dx∫
I =
2
2
16–
2 9
3 +16169– log + –
2 9
xx
c
x x
I = 2 23 16 8 16– – log + – +
2 9 3 9
xx x x c
BA – 5
Q-1) sin x dx∫∫∫∫Ans. Let I = sin x dx∫
Put x = t
∴∴∴∴1
x xdx = dt
∴∴∴∴ dx = 2t dt
∴∴∴∴ I = 2 sint t dt∫= 2 sint t dt∫= 2 sin – sin
dtt t dt t dt dt
dt
∫ ∫ ∫ ∫
= ( )2 – cot + cos +t t t dt c ∫
= [ ]2 – cos +sin +t t t c
I = –2 cos + 2sin +x x x c
Q-2) (((( ))))log
1+ log
xdx
x∫∫∫∫Ans. Let I =
( )
log
1+ log
xdx
x∫Put log x = t
∴∴∴∴ x = et
∴∴∴∴ dx = et dt
∴∴∴∴ I = ( )2
t
1+
te dt
t∫
=( )
( )2
t+1 –1
1+
te dt
t
∫
Indefinite Integration
78 Mahesh Tutorials Science
=( ) ( )
2 2
t +1 1–
1+ 1+
te dt
t t
∫
=( )
1 1–
1+ 1+e dt
t t
∫ 2
t
=( )
2
t 1 1–
1+ 1+e dt
t t
∫
Now, ( )2
1 –1=
1+ 1+
d
dt t t
∴∴∴∴ I =t 1 1
+1+ 1+
de dt
t dt t
∫
=t 1. +1+
e ct
∴∴∴∴ I = +1+ log
xc
x
Q-2) (((( ))))log
1 log
xdx
x++++∫∫∫∫
Ans. Let I = ( )
log
1+ log
xdx
x∫Put log x = t
∴∴∴∴ x = et
∴∴∴∴ dx = et dt
∴∴∴∴ I =( )
2
t
1+
te dt
t∫
=( )
( )2
t+1 –1
1+
te dt
t
∫
=( ) ( )
2 2
t +1 1–
1+ 1+
te dt
t t
∫
=( )
2
t +1 1–
1+ 1+
te dt
t t
∫
=( )
2
t 1 1–
1+ 1+e dt
t t
∫Now,
( )2
1 –1=
1+ 1+
d
dt t t
∴∴∴∴ I =t 1 1
+1+ 1+
de dt
t dt t
∫
=t 1. +1+
e ct
∴∴∴∴ I = +1+ log
xc
x
Q-3)3 + 4 sin
dx
x∫∫∫∫Ans. Let I =
3+ 4sin
dx
x∫Put tan
2
x = t sin x = 2
2
1+
t
t
dx = 2
2
1+
dt
t
I =( )2
2
1 2
2 1+3 + 41+
dt
t t
t
∫
= 23+3 +8
xdt
t t∫=
2
2 1
83+ +13
dtt
t∫
=2
2 1
4 16 73+ 2 + –
3 9 9t t
∫
= 22
2 1
3 4 7+ –3 3
t
∫
=2 1 + 4/3 – 7 /3
log +3 + 4/3 + 7 /372
3
tc
t
I =3tan + 4 – 7
1 2log +7 + 4/3+ 7 /3
x
ct
Q-4)–1tan xdx∫∫∫∫
Ans. I =–1tan x dx∫
=–11.tan x dx∫
Mahesh Tutorials Science 79
Indefinite Integration
=–1 –1tan – tan
dx dx x dx dx
dx
∫ ∫ ∫
=–1
2
1tan –
1+x x x dx
x∫=
–1
2
1 2tan –
2 1+
xx x dx
x∫I = ( )–1 21
tan – log 1 +2
x x x c+
Q-5)sin + sin2
dx
x x∫∫∫∫Ans. Let
I =sin +sin2
dx
x x∫=
sin +2sin cos
dx
x x x∫= ( )sin 1+ 2cos
dx
x x∫
= ( )
sin
sin + 1+2cos
x dx
x x∫ 2
= ( ) ( ) ( )sin
1 – cos 1+ cos 1+2cos
x dx
x x x∫Put cos x = t
∴∴∴∴ –sin x dx = dt
∴∴∴∴ sin x dx = –dt
∴∴∴∴ I = ( ) ( ) ( )–
1– 1+ 1+2
dt
t t t∫
=( ) ( ) ( )
–1– 1+ 1+2
dt
t t t∫
Let ( ) ( ) ( )1
1– 1+ 1+2t t t
= + +1 – 1+ 1+2
A B C
t t t
∴∴∴∴ 1 = A(1 + t)(1 + 2t) + B(1 – t) (1 + 2t) +
C(1 – t) (1 + t)
Putting 1 – t = 0,
i.e., t = 1, we get,
1 = A(2)(3) + B(0)(3) + C(0)(2)
∴∴∴∴ A = 1
6
Putting 1 + t = 0, t = –1, we get,
1 = A(0)(–1) + B(2)(–1) + C(2)(0)
∴∴∴∴ B = 1
–2
Putting 1 + 2t = 0, i.e., t = 1
–2, we get,
1 = ( ) ( )3 1
0 02 2
A B C
+ +
∴∴∴∴ C = 4
3
∴∴∴∴ ( ) ( ) ( )1
1– 1+ 1+2t t t
=
1 –1 4
6 2 3+ +
1– 1+ 1+2t t t
∴∴∴∴ I =
1 –1 4
6 2 3– + +
1– 1+ 1+2dt
t t t
∫
=1 1 1 1 4 1
– + –6 1 – 2 1+ 3 1+2
dt dt dtt t t∫ ∫ ∫
=log 1 –1
– .6 –1
t
log 1+21 4+ log 1+ – . +2 3 2
tt c
=1 1log 1 – cos + log 1+ cos
6 2x x
2– log 1+2cos + .3
x c
Indefinite Integration
80 Mahesh Tutorials Science
Q-2) secn x . tan x
Ans. I = sec .tanx x dx∫ n
= sec .sec tanx x x∫ –1n
Put sec x = y
∴∴∴∴ sec x tan x dx = dy
∴∴∴∴ I =1
= + = +–1+1
yy dy c y c
n n∫–1+1
–1n
n n
∴∴∴∴ I =1
nsecn x + c
Q-3)cos2 – cos2
cos cos
x
x
αααα
αααα⋅⋅⋅⋅
Ans. I =cos2 – cos2
cos .cos
xdx
x
α
α∫
=( ) ( )2cos –1 – 2cos –1
cos cos
xdx
x
α
α⋅∫2 2
=2cos –1 – 2cos +1
cos cos
xdx
x
α
α⋅∫2 2
=2cos 2cos
–cos cos cos cos
xdx
x x
α
α α
⋅ ⋅ ∫2 2
=cos cos
2 – 2cos cos cos cos
xdx dx
x x
α
α α⋅ ⋅∫ ∫2 2
= 2sec cos – 2cos secx dx x dxα∫ ∫= 2sec α sin x – 2 cos α sec + tan +x x c
∴∴∴∴ I =2
cosαsin x – 2 cos α + log
sec + tan +x x c
Q-4) (((( )))) (((( ))))2 3 7 3.cos sinx x x dx∫∫∫∫
Ans. I = ( ) ( )cos sinx x x dx∫ 2 3 7 3
Put sin x3 = y
∴∴∴∴ 3x2 cos x3 dx = dy
∴∴∴∴ x2 cos x3 dx = 1
3 dy
∴∴∴∴ I =
+
1 1 1= = +
73 3 3+1
2
yy dy y dy c∫ ∫
717 2
7 2
=1 2
+3 9
y c9
2× ×× ×× ×× ×
∴∴∴∴ I = ( )2sin +
27x c
93 2
AA–2
Q-1) (((( ))))–1cot sec3 tan3 dθ θ θθ θ θθ θ θθ θ θ++++∫∫∫∫Ans.
1 sin3 1+ sin3sec3 + tan3 = + =
cos3 cos3 cos3
θ θθ θ
θ θ θ
=
1 – cos2+3
sin2+3
π
θ
π
θ
ADVANCED ASSIGNMENTS (AA) :
AA–1
Q-1) sin2
sin – sin + +6 6
x
x x cπ ππ ππ ππ π
∫∫∫∫
Ans. I = ( ) ( )
sin2
sin – 6 sin + 6
xdx
x xπ π∫
=
sin – + +6 6
sin – + +6 6
x x
x x
π π
π π
∫
=
sin – cos +6 6
+ cos – sin +6 6
sin – sin +6 6
x x
x x
dx
x x
π π
π π
π π
∫
= cot + + cot –6 6
x x dxπ π
∫
= log sin + + log sin – +6 6
x x cπ π
Mahesh Tutorials Science 81
Indefinite Integration
=
32sin +
4 2
3 32sin + cos +
4 2 4 2
π θ
π θ π θ
2
=3
tan +4 2
π θ
∴∴∴∴ I =3
cot tan +4 2
dπ θ
θ
∫ –1
=3
cot cot – –2 4 2
dπ π θ
θ
∫ –1
=3
–4 2
dπ θ
θ ∫
=3
– +4 4
cπ θ
θ2
Q-2) sin
sin3
xdx
x∫∫∫∫
Ans. Let I =sin
sin3
xdx
x∫=
sin
3sin – 4sin
xdx
x x∫ 3
=1
3 – 4sindx
x∫ 2
Divide numerator and denominator by sin2x
I =cosec
3cosec – 4
xdx
x∫2
2
=cosec
3cot –1
xdx
x∫2
2
Put cot x = 1 ∴∴∴∴ cosec2x dx = – dt
I =1
–3 –1
dtt∫ 2
=
1 1–
13–3
dt
t∫ 2
=1 1
–3 1
–3
dt
t
∫ 2
2
=
1–
1 1 3– log +
13 1 +233
t
c
t
××××
=1 3 –1
– log +2 3 3 +1
tc
t
∴∴∴∴sin
sin3
xdx
x∫ = 1 3 cot –1
– log +2 3 3 cot +1
xc
x
Q-3) 2
1
cos2 + 3sindx
x x∫∫∫∫
Ans. Let I =1
cos2 +3sindx
x x∫ 2
=1
1– 2sin +3sindx
x x∫ 2 2
=1
1+sindx
x∫ 2
Dividing both numerator and denominator by
I =sec
sec + tan
x dx
x x∫2
2 2
=sec
1+ tan + tan
x dx
x x∫2
2 2
=sec
2tan +1
x dx
x∫2
2
Put tan x = t ∴∴∴∴ sec2x dx = dt
∴∴∴∴ I =1 1 1
=22 +1 1
+2
dt dtt
t
∫ ∫2 2
2
=1 1
tan +12 1
22
tc
–1××××
= ( )1tan 2 tan +
2x c–1
Indefinite Integration
82 Mahesh Tutorials Science
Q-4) 2 + sin2
1+ cos2
x xe dx
x
∫∫∫∫
Ans. Let I =2+sin2
1+cos2
xe dx
x
∫ x
=2+2sin cos
2cos
x xe dx
x
⋅ ∫ 2
x
=2 2sin cos
+2cos 2cos
x xe dx
x x
⋅ ∫ 2 2
x
= sec + tane x x dx ∫ 2x
= tan +sece x x dx ∫ 2x
∴∴∴∴2+sin2
1+cos2
xe dx
x
∫ x = ex tan x + c
Q-5) 2 – sin2
1 – cos2
x xe
x
Ans. I =2 – sin2
1– cos2
xe dx
x
∫ x
=2 – 2sin cos
2sin
x xe dx
x
∫ 2
x
=1 sin cos
–sin sin
x xe dx
x x
∫ 2 2
x
= ( )cosec – cote x x dx∫ 2x
= (–1) ( )cot – cosece x x dx∫ 2x
Let f (x ) = cos x
∴∴∴∴ f ′′′′(x ) = cosec2x
∴∴∴∴ I = ( ) ( ) ( )+ = +e f x f x dx e f x c ∫ x x
= – ex cot x + c
∴∴∴∴ I = (– 1) ex cot x + c
AA–3
Q-1) (((( ))))
2
2
log – 1
1+ log
xdx
x
∫∫∫∫
Ans. I =( )
log –1
1+ log
xdx
x
∫
2
2
Put log x = t
∴∴∴∴ x = et
∴∴∴∴ dx = et dt
∴∴∴∴ I =–1
1+
te dt
t
∫
2
2
t
=( )
– 2 +1
1+
t te dt
t
∫2
22
t
=
( )
( ) ( )
1+ 2–
1+ 1+
t te dt
t t
∫2
2 22 2
t
= ( )
1 2–
1+ 1+
te dt
t t
∫ 2 22
t
Now, 1
1+
d
dt t
2 = ( )1+d
tdt
–12
= (–1) (1 + t2)–1–1d
dx(1+t2)
= (–1) (1+ t2)–2 (0 + 2t)
= ( )
– 2
1+
t
t2
2
∴∴∴∴ I =1 1
+1+ 1+
de dt
dtt t
∫ 2 2
t
=1
+1+
e ct
2
t
=( )
1+
1+ logx c
x
∴∴∴∴ I =( )
+1+ log
xc
x2
Mahesh Tutorials Science 83
Indefinite Integration
Q-2) 1
cos + sin2dx
x x
∫∫∫∫
Ans. I =1
cos +sin2dx
x x
∫
=1
cos + 2sin cosdx
x x x
∫
= ( )
1
cos 1+ 2sindx
x x∫Put sin x = t
∴∴∴∴ cosx dx = dt
∴∴∴∴ dx =cos
dt
x
∴∴∴∴ I =( )cos
cos 1+2sin
dt
x
x x∫
=1+2sin
dt
x∫
= ( ) ( )1 – sin 1+ 2sin
dt
x x∫ 2
= ( ) ( ) ( )1+sin 1 – sin 1+ 2sin
dt
x x x∫
= ( ) ( ) ( )1+ 1 – 1+ 2
dt
t t t∫
Let ( ) ( ) ( )
1
1+ 1 – 1+ 2t t t = + +
1+ 1 – 1+ 2
A B c
t t t
=
( ) ( ) ( ) ( )
( ) ( )
( ) ( ) ( )
1 – 1+ 2 + 1+ 1+ 2
+ 1+ 1 –
1+ 1 – 1+ 2
A t t B t t
C t t
t t t
∴∴∴∴ 1 = A (1 – t ) (1 + 2t ) + B (1 + t ) (1 + 2t )
+ C (1 + t ) (1 – t )
Putting t = – 1, we get
1 = A (1 + 1) (1 – 2) + B (0) + C (0)
∴∴∴∴ A = –1
2
Putting t = 1, we get
1 = A(0) + B(1 + 1) (1 + 2) + C(0)
∴∴∴∴ B = 1
6
Putting t = 1
2, we get
A (0) + B (0) + C1 1
1 – 1+2 2
∴∴∴∴ C = 4
3
∴∴∴∴( ) ( ) ( )
1
1+ 1 – 1+ 2t t t=
–1 1 4
2 6 3+ +1+ 1 – 1+ 2t t t
∴∴∴∴ I =
–1 1 4
2 6 3+ +1+ 1 – 1+2
dtt t t
∫
=1 1 4
+ +2 1+ 6 1 – 3 1+2
dt dt dt
t t t∫ ∫ ∫=
log 1 –1 1log 1+ +
2 6 –1
tt
log 1+24+ +3 2
tc
∴∴∴∴ I = –1 1log 1+ sin – log 1 – sin
2 6x x
2+ log 1+ 2sin +3
x c
Q-3) (((( )))) (((( ))))2
3 – 2
+1 + 3
xdx
x x∫∫∫∫
Ans. I =( ) ( )
3 – 2
+1 + 3
xdx
x x∫ 2
( ) ( )
3 – 2
+1 +3
x
x x2 =
( ) ( ) ( )+ +
+1 +1 + 3
A B C
x x x
3x – 2 = A(x + 1) (x + 3) + B (x + 3)
+ C(x + 1)2 ...(i)
Put x = – 1 in (i), we get
– 5 = B(2) ⇒⇒⇒⇒ B = –5
2
Put x = – 3 in (1), we get
–11 = 4 C C = –11
4
Comparing co-efficient of x2
A + C = 0
Indefinite Integration
84 Mahesh Tutorials Science
A –11
4= 0 ⇒⇒⇒⇒ A =
11
4
∴∴∴∴ I =( ) ( ) ( )
–511 –11
24 4+ ++1 +3+1
dxx xx
∫ 2
= ( )
11 1 5 1 11– –
4 +1 2 4+1dx dx
x x∫ ∫ 2
( )
1
+3dx
x∫
= ( )
11 5 1log +1 +
4 2 +1x
x
11– log + 3 +4
x c
=( )
11 +1 5log + +
4 + 3 2 +1
xc
x x
Q-4)
2
3 2
5 + 20 +
+ 2 +
x x cdx
x x x∫∫∫∫
Ans. Let I =5 + 20 + 6
+ 2 +
x xdx
x x x∫2
3 2
= ( )5 + 20 +6
+ 2 +1
x xdx
x x x∫2
2
=( )
5 + 20 + 6
+1
x xdx
x x∫2
2
Let ( )
5 +20 +6
+1
x x
x x
2
2 = ( )
+ ++1 +1
A B C
x x x2
∴∴∴∴ 5x2 + 20x + 6 = A (x + 1)2 + Bx (x + 1) + Cx
Put x = 0, we get,
0 + 0 + 6 = A (1) + B (0)(1) + C (0) ∴∴∴∴ A = 6
Put x + 1 = 0, i.e., x = – 1, we get,
5(1) + 20(– 1) + 6 = A (0) + B (–1) (0) + C (–1)
∴∴∴∴ – 9 = – C ∴∴∴∴ C = 9
Put x = 1, we get,
5(1) + 20(1) + 6 = A (4) + B (1)(2) + C (1)
But A = 6 and C = 9
∴∴∴∴ 31 = 24 + 2B + 9 ∴ B = – 1
∴∴∴∴( ) ( )
5 +20 +6 6 1 9= – +
+1+1 +1
x x
x xx x x
2
2 2
∴ I =( )
6 1 9– +
+1 +1dx
x x x
∫ 2
= ( )1 1
6 – +9 +1+1
dx dx x dxx x∫ ∫ ∫
–2
=( )+1
6 log – log +1 + 9 +–1
xx x c⋅
–1
=9
6 log – log +1 – ++1
x x cx
Q-5) (((( )))) (((( ))))tan – tan + .tan2x x xdxα αα αα αα α∫∫∫∫
Ans. Let I = ( ) ( )tan – tan + .tan2x x x dxα α∫Let tan 2x = tan [(x + α) + (x – α)
=( ) ( )
( ) ( )
tan + + tan –
1 – tan + tan –
x x
x x
α α
α α⋅
∴∴∴∴ tan 2x [1 – tan (x + α) tan (x – α)]
= tan (x + α) + tan (x – α)
∴∴∴∴ tan 2x – tan 2x tan (x + α) tan (x – α)
= tan (x + α) + tan (x – α)
∴∴∴∴ tan 2x – tan (x + α) – tan (x – α)
= tan 2x . tan (x + α) . tan (x – α)
∴∴∴∴ I =( )
( )
tan2 – tan +
– tan –
x xdx
x
α
α
∫
= ( )1log sec2 – log sec +
2x x α
( )– log sec – +x cα
AA–4
Q-1)
–1 –1
–1 –1
sin – cos
sin + cos
x xdx
x x∫∫∫∫
Ans. I =sin – cos
sin + cos
x xdx
x x∫–1 –1
–1 –1
=
sin – – sin2
2
x x
dx
π
π
∫
–1 –1
=2
2sin –2
x dxπ
π
∫ –1
Mahesh Tutorials Science 85
Indefinite Integration
=4
sin – 1x dx dxπ ∫ ∫–1
=4
sin – +x dx x cπ ∫ –1
=4
– +I x cπ
1
Now, I1= sin x dx∫ –1
Put x = sin2 θθθθ
dx = sin2θθθθ dθθθθ
∴∴∴∴ dx = 2 sin θθθθ . cos θθθθ dθθθθ
I = sin sin sin2 dθ θ θ ∫ –1 2
I = sin2 dθ θ θ⋅∫=
cos2 1– + cos2
2 2d
θθ θ θ∫
= ( )1 1
– 1 – 2sin + sin 1 – sin2 2
θ θ θ θ2 2
= ( )1 1
– 1 – 2 sin + 1 –2 2
x x x x⋅–1...(ii)
From (i) and (ii), we get
I = ( )4 1 1
– 1– 2 sin + – – +2 2
x x x x x cπ
–1 2
= ( )2
– – 1 – 2 sin – +x x x x x cπ
2 –1
Q-2) (((( ))))(((( ))))
2
1log log +
logx dx
x
∫∫∫∫
Ans. Let I = ( )( )
1log log +
logx dx
x
∫ 2
Put log x = t ∴∴∴∴ x = et
∴∴∴∴ dx = et dt
∴∴∴∴ I =1
log +e t e dtt
∫ 2
t t
=1 1 1
log + – +e t dtt t t
∫ 2
t
=1 1 1
log + + – +e t e dtt t t
∫ 2
t t
=1 1 1
log + – –e t dt e dtt t t
∫ ∫ 2
t t
= I1 – I
2
In I1, Put f (t ) = log t. Then f ′′′′(t ) = (1/t )
∴∴∴∴ I1= ( ) ( )+e f t f t dt′ ∫ t
= et f (t ) = et log t
In I2, Put g (t) = (1/t). Then g′′′′(t) = – (1/t2)
∴∴∴∴ I2= ( ) ( )+e g t g t dt′ ∫ t
= et g (t) = et . (1/t).
∴∴∴∴ I = et log t – e
t
t
+ c
= x log (log x) –log
x
x+ c.
Q-3) tan
1 – sin
xdx
x∫∫∫∫
Ans. Let I =tan
1 – sin
xdx
x∫
= ( )
sin
cos 1 – sin
xdx
x x∫
= ( )
sin .cos
cos 1 – sin
x xdx
x x∫ 2
=( ) ( )
sin .cos
1 – sin 1 sin
x xdx
x x+∫ 2
Put sin x = t
∴∴∴∴ cos x dx = dt
I =( ) ( )1 – 1+
tdt
t t∫ 2
We express
( ) ( )1 – 1+
t
t t2
=( ) ( )
+ +1 – 1+1 –
A B C
t tt2
t = A(1 – t) (1 + t) + B(1 + t) + C(1 – t)2
t = (1 + t) [A (1 – t) + B] + C (1 – t)2
...(i)
Put t = –1 in equation (i), we get
–1 = 4C ⇒⇒⇒⇒ C = –1
4
Indefinite Integration
86 Mahesh Tutorials Science
Put t = 1 in equation (i), we get
1 = 2 [0 + B] ⇒⇒⇒⇒ B = 1
2
Put t = 0 in equation (i), we get
0 = 1[A + B] + C
0 =1 1
+ –2 4
A
∴∴∴∴ A =1–4
∴∴∴∴( ) ( )1 – 1+
t
t t2
=( )
1–1 –1
24 4+ +1 – 11 –t tt +2
∴∴∴∴ I = ( )1 1 1 1 1
– + 1 – –4 1 – 2 4 1+
dt t dt dtt t∫ ∫ ∫
–2
= ( )
1 1 1log 1 – + – log 1+ +
4 2 1 – 4t t c
t
=( )
1 1 – 1log + +
4 1+ 2 1 –
tc
t t
=( )
1 1 – sin 1log + +
4 1 sin 2 1 – sin
xc
x x+
Q-4) (((( ))))log 1 cos – tan2
xx x dx
++++
∫∫∫∫
Ans. I = ( )log 1+ cos – tan2
xx x dx
∫
= ( )log 1+ cos – tan2
xx dx x dx∫ ∫ ...(i)
= ( )log 1+ cos x dx∫= ( ) ( )log 1+ cos 1x dx∫= ( ) ( )log 1+ cos
dx x dxdx∫
= ( ) ( ) ( )log 1+ cos – log 1+ cos .d
x x x x dxdx
∫
= ( )2sin cos
2 2log 1+ cos + .
2cos2
x x
x x x dxx∫ 2
= ( )log 1+ cos + .tan +2
xx x x dx c∫
From (i). we get
I = x log (1 + cos x)
+ tan + – tan2 2
x xx dx c x dx∫ ∫
∴∴∴∴ I = x log (1 + cos x) + c
Q-5) cos2
sindx
θθθθ
θθθθ∫∫∫∫
Ans. I =cos2
sind
θθ
θ∫ =1 – 2sin
sind
θθ
θ∫2
=1 – 2sin
sind
θθ
θ∫2
2=
1– 2
sindθ
θ∫ 2
= cos – 2ec dθ θ∫ 2= ( )1+ cot – 2dθ θ∫ 2
= cot –1dθ θ∫ 2=
cot –1
cot –1
θ
θ∫2
2
=cosec
– 2cot –1 cot –1
dd
θ θθ
θ θ∫ ∫2
2 2
= I1 – 2I
2...(i)
Now,
I1=
cosec
cot –1d
θθ
θ∫2
2
Put cot θθθθ = t
∴∴∴∴ – cosec2 θθθθ dθθθθ = dt
∴∴∴∴ cosec2 θθθθ dθθθθ = – dt
∴∴∴∴ I1=
––
–1 –1
dt dt
t t
=∫ ∫2 2
= – log + –1 +t t c2
1
= – log cot + cot –1 + cθ θ2 1 ...(ii)
Also, I2=
cot –1 cos–1
sin
d dθ θ
θ θ
θ
=∫ ∫2 2
2
=sin
cos – sind
θθ
θ θ∫ 2 2
Mahesh Tutorials Science 87
Indefinite Integration
I2=
sin
2cos –1d
θθ
θ∫ 2
Put cos θθθθ = u
∴∴∴∴ – sin θθθθ dθθθθ = du
∴∴∴∴ sin θθθθ dθθθθ = –du
∴∴∴∴ I2
=–
2 –1
du
u∫ 2=
––
12 –
2
du
u
∫2
=1
–2 1
–2
du
u∫ 2
=1 1
– log + – +2 2
u u c2
2
I2
=1 1
– log cos + cos – +2 2
cθ θ2
2 ...(iii)
From (i), (ii) and (iii), we get
I = – log cot + cot –1 + cθ θ21
1 1+2 log cos + cos – +
22cθ θ
2
2
– log cot + cot –1θ θ2
1+ 2 log cos + cos – +
2cθ θ2
AA–5
Q-1) cos 7 – cos8
1 2cos5
x xdx
x++++∫∫∫∫Ans. I =
cos7 – cos8
1 2cos5
x xdx
x+∫
=
7 – 8 8 – 72sin sin
2 2
51 2 1 – sin
2
x x x x
x
+
∫ 2
=
15sin .sin
2 225
3 – 4sin2
x x
dxx∫ 2
=
15 5sin .sin .sin
2 2 225 5
3sin – 4sin2 2
x x x
dxx x∫ 3
=
15 5sin .sin .sin
2 2 225
sin32
x x x
dxx
∫
=
15 5sin .sin .sin
2 2 2215
sin2
x x x
dxx∫
=5
2sin .2 2
x xdx∫
=5 5
cos – – cos +2 2 2 2
x x x xdx
∫
= ( )cos2 – cos3x x dx∫
= cos2 – cos3x dx x dx∫ ∫∴∴∴∴ I =
sin2 sin3– +
2 3
x xc
Q-2) 2 2 –
x
x x
edx
a e b e++++∫∫∫∫
Ans. Let I =+
edx
a e b e∫ 2 2 –
x
x x
=
+
edx
ba e
e
∫ 22
x
x
x
=+
edx
a e b∫ 2 2
x
2x
Put ex = t ∴∴∴∴ ex dx = dt.
∴∴∴∴ I =1
+dt
a t b∫ 2 2 2
=1 1
+
dta b
ta
∫ 2
2
Indefinite Integration
88 Mahesh Tutorials Science
=1log + + +
bt t c
a a
2
=1log + + +
be e c
a a
2
2x x
Q-3) 3 2sin2 4 cos 2
dx
x x+ ++ ++ ++ +∫∫∫∫Ans. Let I =
3+2sin2 + 4cos2
dx
x x∫Put tan x = t ∴∴∴∴ x = tan–1 t
∴∴∴∴ dx =1+
dt
t 2 and
sin 2x =2
,1+
t
t2 cos 2x = 1 –
1+
t
t
2
2
∴∴∴∴ I =1
.1+2 1 –
3 + 2 + 41+ 1+
dt
tt t
t t
∫ 22
2 2
=( ) ( )
1+.1+3 1+ + 4 + 4 1 –
t dt
tt t t∫2
22 2
=1
7 + 4 –dt
t t∫ 2
= ( )1
7 – – 4 + 4 + 4dt
t t∫ 2
=( ) ( )
1
11 – – 2
dt
t∫ 2 2
=1 11 + – 2
log2 11 11 – 2
tc
t+
+
=1 11 + tan – 2
log +2 11 11 – tan + 2
xc
x
Q-4) (((( )))){{{{ }}}}2 2
4
1 log 1 – 2 logx x x
dxx
+ ++ ++ ++ +
∫∫∫∫
Ans. Let I = ( ){ }+1 log +1 – 2logx x x
dxx∫
2 2
4
= ( )+1 1
log +1 – log . .x
x x dxx x
∫
22 2
3
=+1 +1 1
log . .x x
dxx x x
∫
2 2
2 2 3
=1 1 1
log 1+ . 1+ . dxx x x
∫ 2 2 3
Put 1
1 tx
+ =2
∴∴∴∴ –2x–3 dx = dt
∴∴∴∴1dx
x3=
–
2
dt
∴∴∴∴ I = ( )( )–
log . .2
dtt t∫
1
2
= ( )1
– log .2
t t dt∫1
2
= ( ) ( )1
– log – log2
dt t dt t t dt dt
dt
∫ ∫ ∫
1 1
2 2
= ( )1 1
– log . – .2 3/2 3/2
t tt dt
t
∫3 3
2 2
=1 2 2
– .log –2 3 3
t t t dt ∫
3 1
2 2
=1 2
– log – +2 3 3/2
tt t c
×
33 22
=1 2
– log –3 3t t c
+
3
2
=1 1 1 2
– 1+ log 1+ – +3 3
cx x
3
2
2 2
Q-5) (((( ))))
2
1 1–
log logdx
x x
∫∫∫∫
Ans. Let I = ( )
1 1–
log logdx
x x
∫ 2
Put log x = t ∴ ∴ ∴ ∴ x = et
∴∴∴∴ dx = et dt
∴∴∴∴ I =1 1– .e dt
t t
∫ 2
t
=1 1–e dt
t t
∫ 2
t
Mahesh Tutorials Science 89
Indefinite Integration
AA–6
Q-1) (((( ))))2
log 2 logxe
x x x dxx
++++ ∫∫∫∫
Ans. I = ( )log + 2loge
x x x dxx ∫
2x
= ( )2log
log +x
e x dxx
∫
2x
Put f(x ) = (log x)2
∴∴∴∴ f ′(x) = ( )logd
xdx
2 = 2 (log x). ( )log
dx
dx
=2log x
x
∴∴∴∴ I = ( ) ( )+e f x f x dx′ ∫ x
= ex. f(x) + c = ex. (log x)2 + c.
Q-2) 1 –
1
xdx
x++++∫∫∫∫
Ans. Let I =1 –
1+
xdx
x∫Put =x t ∴∴∴∴ x = t ∴∴∴∴ dx = 2t dt
I =1 –
21+
tt dt
t∫
=( ) ( )
( ) ( )
1 – 1 –2
1+ 1 –
t tt dt
t t∫
=( )1 –
21 –
tt dt
t∫2
2
=( )1 –
21 –
t tdt
t∫ 2
=–
21 –
t tdt
t∫2
2
Q-3) (((( ))))2
2
3 3
xdx
x x
++++
+ ++ ++ ++ +∫∫∫∫
Ans. I = ( )
+ 2
+3 +3
xdx
x x∫ 2
Let x + 2 = a (2x + 3) + b
x + 2 = 2ax + 3a + b
Comparing co-efficents
2a = 1 3a + b = 2 ∴∴∴∴ 3
2+ b = 2
a =1
2 b = 2 –
–3
2 ; ∴∴∴∴ b =
1
2
I =
( )1 12 + 3 +
2 23 9 3
+ 2 + +2 4 4
x
dx
x x
∫ 2
=
3 1+
2 2
3 3+ +2 4
x
dx
x
∫ 2
Let (f )t =1
t . ∴ ∴ ∴ ∴ f ′′′′(t) =
–1
t2
∴∴∴∴ I = ( ) ( )e f t f t dt′+ ∫ t
= ( )1
.e f t c e ct
+ = × +∫ t t
= +log
xc
x
= 2 –1 – 1 –
t tdt
t t
∫
2
2 2
=1 2 –1+1
2 –2 1 – 1 –
t tdt dt
t t
∫ ∫
2
2 2
= ( )1 12 –2 1 – –
2 1 –t dt
t
∫2
2
–1–
1 –
tdt
t
∫
2
2
= 2 – 1 – + cos + 1 –t t dt
2 –1 2
= 2 – 1– + cos + 1–2
tt t
2 –1 2
1+ sin2 1
t
–1
= –2 1 – +2cos + 1 – + sint t t t t2 –1 2 –1
I = –2 1 – + 2cosx x–1
+ 1– +sin +t x x c–1
I = –2 1 – + cos + 1 – +x x x x c–1
Indefinite Integration
90 Mahesh Tutorials Science
=
3 1+2 2 +3 3+ +2 4
x
dx
x
∫ 2
1
2
3 3+ +2 2
dx
x
∫ 22
=
32 +
1 2+
2 3 3+ +2 4
x
dx
x
∫ 2
1 1
2 3 3+ +2 2
dx
x
∫ 22
=1 3 3log + +2 2 4
x
2
3+
1 2+ tan3 3
2 2
x
–1
= 1log + 3 + 32
x x2
2 2 + 3+ tan +3 3
xc
–1
Q-5) (((( ))))
3
2
1
xdx
x
++++
++++∫∫∫∫
Ans. Let I =( )
+ 2
+1
xdx
x∫ 3
Let ( )
2
1
x
x
+
+3 =
( ) ( )+ +
+1 +1 +1
A B C
x x x2 3
∴∴∴∴ x + 2 = A (x + 1)2 + B(x + 1) + C ...(i)
Put x + 1 = 0, i.e., x = –1, we get,
–1 + 2 = A (0) + B (0) + C ∴∴∴∴ C = 1
Put x = 0 in (i), we get,
0 + 2 = A (1) + B (1) + C
∴∴∴∴ 2 = A + B + 1
∴∴∴∴ A + B = 1 ...(ii)
Put x = 1 in (i), we get,
1 + 2 = A(4) + B(2) + C
∴∴∴∴ 3 = 4A + 2B + 1
∴∴∴∴ 4A + 2B = 2 ∴∴∴∴ 2A + B = 1 ...(iii)
Subtracting (ii) from (iii), we get,
A = 0
∴∴∴∴ from (ii), 0 + B = 1 ∴∴∴∴ B = 1
∴ ∴ ∴ ∴ ( )
+2
+1
x
x3=
( ) ( )
0 1 1+ +
+1 +1 +1x x x2 3
=( ) ( )
1 1+
+1 +1x x2 3
∴∴∴∴ I =( ) ( )
1 1+
+1 +1dx
x x
∫ 2 3
= ( ) ( )+1 + +1x dx x dx∫ ∫–2 –3
=( ) ( )+1 +1
+ +–1 –2
x xc
–1 –2
=( )
–1 1– +
+1 2 +1c
x x2
AA–7
Q-1) 1
cos cos2dx
x x∫∫∫∫
Ans. Let I = ∫1
cos cos2dx
x x
= ∫ 2
2
sec
1 – tan
1+ tan
xdx
x
x
= ∫ 2
××××sec sec
1 – tan
x xdx
x
= ∫2
2
sec
1 – tan
xdx
x
Put tanx = t ∴∴∴∴ sec2x dx = dt
∴∴∴∴ I = ∫ 2
1
1 – tdt = sin–1 t + c
= sin–1 (tanx) + c
Mahesh Tutorials Science 91
Indefinite Integration
Q-2) (((( )))) (((( ))))–2
log log + logx x
Ans. I = ( )( )
∫ 2
1log log +
logx dx
x
Put logx = t
∴∴∴∴ x = et ∴∴∴∴ dx = etdt
∴∴∴∴ I = ∫ t
2
1log +t e dt
t
= ∫ t
2
1 1 1log + – +e t dt
t t t
= ∫ t
2
1 1 1log – + +e t dt
t t t
Put f(t) = logt – 1
t
∴∴∴∴ f′′′′(t) = 2
1 1+t t
∴∴∴∴ I = ( ) ( )′ ∫ t +e f t f t dt = etf(t) + c
= ( )
logx 1log log – +
loge x c
x
∴∴∴∴ I = x log (log x) – +log
xc
x
Q-3) tan∫∫∫∫ xdx
Ans. Let I = tanx dx∫Put tan x = t2
∴∴∴∴ sec2 x dx = 2t dt
dx = 2
2
1+ tan
t dt
x
= 4
2
1+
tdt
t
∴∴∴∴ I =2
4
2
1+
tt dt
t⋅∫
I =
2
4
2
1+
tdt
t∫
=( ) ( )2 2
4
+1 + –1
1+
t tdt
t∫
=
2
4
+1
+1
t
t∫ dt +
2
4
–1
+1
t
t∫ dt
= I1 + I
2...(i)
Here I1
=
2
4
+1
+1
t
t∫ dt
Divide numerator and denominator by t2
=2
2
2
11+
1+
t
tt
∫ dt
=2
2
11+
1– + 2
t
tt
∫ dt
Put1
– =t zt
∴∴∴∴ 2
11+
t
dt = dz
I1=
( )2
2
1
+ 2dz
z∫
=–11
tan +2 2
zc
=–1
1
1–
1tan +
2 2
tt c
=
2–1
1
1 –1tan +
2 2
tc
t
⋅
=–1
1
1 tan –1tan +
2 2tan
xc
x
...(ii)
Let I2=
2
4
–1
+1
t
t∫ dt
Divide numerator and denominator by t2
∴∴∴∴ I2=
2
2
2
11 –
1+
t
tt
∫ dt
=
2
2
11 –
1+ – 2
t
tt
∫ dt
Indefinite Integration
92 Mahesh Tutorials Science
Put1
+ =tt
θ ∴∴∴∴ 2
11–
t
dt = dθ
∴∴∴∴ I2=
( )2
2
1
– 2
dθ
θ
⋅∫
= 2
1 – 2log +
2 2 + 2c
θ
θ
= 2
1+ – 2
1log +
12 2 + + 2
tt
c
tt
=2
22
1 – 2 +1log +
2 2 + 2 +1
t tc
t t
= 2
1 tan – 2tan +1log +
2 2 tan + 2tan +1
x xc
x x
From (ii) and (iii), equation (i) becomes
I =–11 tan –1
tan +2 2tan
x
x
1 tan – 2tan +1log +
2 2 tan + 2tan +1
x xc
x x
Q-4)1
sin + tandx
x x∫∫∫∫
Ans. I =1
sin + tandx
x x∫
=cos
sin cos +sin
xdx
x x x∫
=( )
cos
sin 1+ cos
xdx
x x∫
=cos
cosec1+ cos
xx dx
x∫
=( )2cos /2 –1
cosec2cos /2
xx dx
x∫2
2
=1
cosec 1 –2cos /2
x dxx
∫ 2
=cosec
cosec –2 cos / 2
xx dx
x∫ 2
=cosec
cosec –1+ cos
xx dx dx
x∫ ∫
= log |cosec x – cot x|
( )
1–sin 1+cos
dxx x∫
Let I1
= ( )
1
sin 1+ cosdx
x x∫
= ( )
sin
sin 1+ cos
xdx
x x∫ 2
= ( ) ( )
sin
1 – cos 1+ cos
xdx
x x∫ 2
Put cos x = t ∴∴∴∴ – sin x dx = dt
I1
=( ) ( )
–
1 – 1+
dt
t t∫ 2
= ( ) ( ) ( )
–1
1 – 1+ 1+dt
t t t∫
=( ) ( )
–1
1 – 1+dx
t t∫ 2
( ) ( )
–1
1 – 1+t t2
=( ) ( ) ( )
+ +1+ 1+ 1 –
A B C
t t t
–1 = A (1 + t) (1 – t) + B (1 – t)
+ C (1 + t)2 (2)
Put t = 1 in (2)
–1 = C(2)2 ⇒⇒⇒⇒ C = –1
4
Put t = –1 in (2)
–1 = B (1 + 1) B = –1
2
Compaing co-efficents of t2
A + C = 0 ∴∴∴∴ A = 1
4
I1
=( ) ( )
1/4 –1/2 –1/4+ +
1+ 1 –1+dt
t tt
∫ 2
=
( )
1 1 –1 1 1 1–
4 1+ t 2 4 1 –1+dt dt dt
tt∫ ∫ ∫2
= ( )
1 1 1log 1+ + + log 1 – +
4 2 1+ 4t t c
t
Mahesh Tutorials Science 93
Indefinite Integration
I1
=( )
1 1log 1+ cos +
4 2 1+ cosx
x
1+ cos 1 – cos +4
x c
I =
1log cosex – cot – log 1+ cos
4x x x
= ( )
–1 1– log 1 – cos +
2 1+ cos 4x c
x
Q-5) (((( ))))2
log 2 log ∫∫∫∫
xe
x x x dxx
+
Ans. Let I = ( )2
x
log + 2loge
x x x dxx ∫
= ( )2x 2log
log +x
e x dxx
∫
Put f (x) = (logx)2
∴∴∴∴ f ′′′′ (x) =
d
dx(logx)2 = ( ) ( )2 log log
dx x
dx⋅
=2log x
x
∴∴∴∴ I = ( ) ( )x +e f x f x dx′ ∫= ex .f (x) + c = ex . (logx)2 + c .
AA–8
Q-1) (((( ))))3
1
sin sin +dx
x x αααα⋅⋅⋅⋅∫∫∫∫
Ans. Let I =( )α⋅∫ 3sin sin +
dxdx
x x
=( )α α⋅∫ 3sin sin cos + cos sin
dx
x x x
=( )α α∫ 4sin cos + cot sin
dx
x x
=α α⋅∫
2cosec d
cos +cot sin
x x
x
Put cos α + cot x.sin α = t
∴∴∴∴ –cosec2x.sin α dx = dt
∴∴∴∴ cosec2x dx = –αsin
dt
∴∴∴∴ I = α
∫
1 –
sin
dt
t=
α∫ ∫ –1 2–1
sint dt
=α
⋅1 2–1
+sin 1 2
tc
Q-2)
3
11
cos
sin
xdx
x∫∫∫∫
Ans. I = ∫3
11
cos
sin
xdx
x
= ⋅∫3
3 8
cos 1
sin sin
xdx
x x
= ⋅∫ 3
4
1cot
sinx dx
x
= ⋅ ⋅∫3
22 2
1 1cot
sin sinx dx
x x
= ⋅ ⋅∫3
2 22cot cosec cosecx x xdx
Put cotx = t
∴∴∴∴ –cosec2x dx = dt
∴∴∴∴ cosec2x dx = –dt
∴∴∴∴ I = ( ) ( )∫3
22t 1+ t –dt = – ∫
3 7
2 2t + t dt
= – ∫ ∫3 7
2 2t – tdt dt
=
3 7+1 +1
2 2– t t– +
3 7+1 +1
2 2
c
=
5 9
2 22 2
– t – t +5 9
c
∴∴∴∴ I =
5 9
2 22 2
– cot – cot +5 9
x x c
Indefinite Integration
94 Mahesh Tutorials Science
Q-3) 2
1
sin cos + 2cosdx
x x x⋅⋅⋅⋅∫∫∫∫
Ans. I =⋅∫ 2
1
sin cos +2cosdx
x x x
Divide by cos2x
I = ∫2sec
tan +2
xdx
x
f(x) = tanx + 2 f′′′′(x) = sec2x
I = log tan +2 +x c
Q-4)
2
– –
x
x x
edx
e e∫∫∫∫
Ans. I = ∫2
–
x
x x–
edx
e e
= ∫2x
x
x
1–
edx
ee
=
( )∫2
2
x
x
x
1 –
edx
e
e
=
( )∫2
2
x x
x1 –
e edx
e
=
( )
⋅
∫2 2
2
x x
x1 –
e edx
e
= ( )∫ 2
x
x1 –
edx
e
Put ex = t ∴∴∴∴ ex dx = dt
I = ∫ 2
1
1 –dx
t
= sin–1 t + c
= ∫2
–
x
x x–
edx
e e
= sin–1 (ex ) + c
Q-5) 1 secx dx∫∫∫∫ +
Ans. I = ∫ 1+sec x dx
= ∫1
1+sec
dxx
= ∫cos +1
cos
xdx
x= ∫
22cos2
cos
x
dxx
= ∫ 2
2 cos2
1 – 2sin2
x
x
Put sin2
x = t
∴∴∴∴
1cos
2 2
x dx = dt
∴∴∴∴ cos2
xdx = 2dt
∴∴∴∴ I = ∫ 2
12
1 – 2t × × × × 2dt
=
∫2
12 2
12 –
2
dt
t
=
∫ 2
2
2 2 1
2 1–
2
dt
t
= –12sin +
1
2
tc
∴∴∴∴ I =
–12sin 2sin +2
xc