Topic Page No.
Theory 01 - 03
Exercise - 1 04 - 11
Exercise - 2 12 - 15
Exercise - 3 16 - 17
Exercise - 4 18
Answer Key 19 - 23
Contents
INDEFINITE INTEGRATION
SyllabusIntegration as the inverse process of differentiation, indefinite integrals of
standard function, integration by parts, integration by the methods of
substitution and partial fractions
Name : ____________________________ Contact No. __________________
ARRIDE LEARNING ONLINE E-LEARNING ACADEMYA-479 indra Vihar, Kota Rajasthan 324005
Contact No. 8033545007
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KEY CONCEPTS
1. DEFINITION :If f & g are function of x such that g’ (x) = f (x) then the function g is called a PRIMITIVE ORANTIDERIVATIVE OR INTEGRAL of f(x) w.r.t. x and is written symbolically as
)x(f}c)x(g{dxdc)x(gdx)x(f =+Û+=ò , where c is called the constant of integration.
2. STANDARD RESULTS:
(i) 1nc)1n(a
)bax(dx)bax(1n
n -¹++
+=++
ò (ii) cbaxna1
baxdx ++=
+ò l
(iii) cea1dxe baxbax += ++ò (iv) c)0a(
naa
p1dxa
qpxqpx +>=
++ò l
(v) ò ++-=+ c)baxcos(a1dx)baxsin( (vi) ò ++=+ c)baxsin(
a1dx)baxcos(
(vii) ò ++=+ c)baxsec(na1dx)baxtan( l (viii) ò ++=+ c)baxsin(n
a1dx)baxcot( l
(ix) ò ++=+ c)baxtan(a1dx)bax(sec2 (x) ò ++-=+ c)baxcot(
a1dx)bax(eccos 2
(xi) c)bax(eccosa1dx)baxcot(.)bax(eccos ++-=++ò
(xii) c)xtanx(secndxxsec ++=ò l OR ò +÷øöç
èæ +p= c
2x
4tanndxxsec l
(xiii) ò +-= c)xcotxec(cosndxxeccos l OR ò += c2xtanndxxeccos l OR )xcotecx(cosn +- l
(xiv) ò +=-
- caxsin
xadx 1
22 (xv) ò +=+
- caxtan
a1
xadx 1
22
(xvi) ò +=-
- caxsec
a1
axxdx 1
22 (xvii) ò úûù
êëé ++=
+
2222
axxnax
dx l
(xviii) ò úûù
êëé -+=
-
2222
axxnax
dx l (xix) ò +-+=
-c
xaxan
a21
xadx
22 l
(xx) ò ++-=
-c
axaxn
a21
axdx
22 l (xxi) ò ++-=- - caxsin
2axa
2xdxxa 1
22222
INDEFINITE INTEGRATION
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(xxii) ò +++=+ - caxsinh
2aax
2xdxax 1
22222 (xxiii) ò +--=- - c
axcosh
2aax
2xdxax 1
22222
(xxiv) ò +-+
= c)bxsinbbxcosa(ba
edxbxsin.e 22
axax
(xxv) ò +++
= c)bxsinbbxcosa(ba
edxbxcos.e 22
axax
3. TECHNIQUES OF INTEGRATION :(i) Substitution or change of independent variable.
Integral I = ò )x(f dx is changed to ò f )t('f))t((f dt, by a suitable substitution x = f (t) provided the
later integral is easier to integrate.
(ii) Integration by part : ò ò ò ò úûù
êëé-= dxv.dxdudxvudxv.u dx where u & v are dif ferentiable func-
tion . Note: While using integration by parts, choose u & v such that
(a) ò dxv is simple & (b) ò ò úûù
êëé dxv.dxdu
dx is simple to integrate.
This is general ly obtained, by keeping the order of u & v as per the order of the letters in ILATE,where; I - Inverse function, L- Logarithmic function,
A- Algebraic function, T- Trigonometric function & E-Exponential function,
(iii) Partial fraction, splitting a bigger fraction into smaller fraction by known methods.
4. INTEGRALS OF THE TYPE :
(i) [ ]ò dx)x('f)x(f nOR ò n)]x(f[
)x('f dx put f(x) = t & proceed.
(ii) òòò ++++++
dxcbxax,cbxax
dx,cbxax
dx 222
Express ax2 + bx + c in the from of perfect square & then apply the standard results.
(iii) òò ++
+++
+ dxcbxax
qpx,dxcbxax
qpx22
Express px + q = A (differential coefficient of denominator ) + B.
(vi) ò +=+ c)x(f.edx)]x('f)x(f[e xx (v) ò +=+ c)x(fxdx)]x('xf)x(f[
(vi) ò Î+
Nn)1x(x
dxn Take xn common & put 1 + x–n = t.
(vii)( )ò Î
+- Nn
1xx
dx
n)1n(
n2, take xn common & put 1 + x–n = tn
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(viii)( )ò +
n/1nn x1x
dx take xn common as x and put 1 + x–n = t.
(ix) ò + xsinbadx
2 OR ò + xcosbadx
2 OR ò ++ xcoscxcosxsinbxsinadx
22
Multiply N’ & D’ by sec2 x & put tan x = t .
(x) ò + xsinbadx
OR ò + xcosbsadx
OR ò ++ xcoscxsinbadx
Hint : Conv er t sines & cosines into thei r respect i v e tangents of ha l f the angles,
put tan t2x
=
(xi) ò ++++ .dx
nxsin.mxcos.cxsin.bxcos.a
l Express Nr º A (Dr) + B dxd
(Dr) + c & proceed.
(xii) ò +++ dx
1Kxx1x24
2OR ò ++
- dx1Kxx
1x24
2where K is any constant.
Hint: Divide Nr & Dr by x2 & proceed.
(xiii) ò ++ qpx)bax(dx
& ò =++++
22 tqpxput;
qpx)cbxax(dx
(xiv) ò =++++
;t1baxput,
rqxpx)bax(dx
2 ò ++ rpx)cax(dx
22 , put x = t1
(xv) ò -ba-x
x dx or ò -ba- )x)(x( ; put x = a cos2 q + b sin2 q
ò b-a-
xx
dx or ò b-a- )x)(x( ; put x = a sec2 q – b tan2 q
ò b-a- )x)(x(dx
; put x – a = t2 or x – b = t2.
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PART - I : OBJECTIVE QUESTIONS
* Marked Questions are having more than one correct option.
Section (A) : Integration Using Standard Integral
A-1. The value of ò a+ )xsin(.xsindx
is equal to
(A) cosec a ln C)xsin(
xsin+
a+ (B) cosec a ln Cxsin
)xsin(+
a+
(C) cosec a ln Cxsec
)xsec(+
a+(D) cosec a ln C
)xsec(xsec
+a+
A-2.1 xdx tan a b
1 sinx 2æ ö= + +ç ÷+ è øò , then
(A) a ,b R4p
= - Î (B) a ,b R4p
= Î (C) 5a ,b R4p
= Î (D) none of these
A-3. If 1(sin2x cos2x) dx sin(2x a) b,2
- = - +ò then
(A) 5a ,b R4p
= Î (B) 5a ,b R4p
= - Î (C) a ,b R4p
= Î (D) none of these
A-4. The value of cos2x dxcosxò is equal to
(A) 2 sin x – l n |sec x + tan x| + C (B) 2 sin x – l n |sec x – tan x| + C(C) 2 sin x + l n |sec x + tan x| + C (D) None of these
Section (B) : Integration Using Substitution
B-1. The value of xa dxxò is equal to
(A) xa Cx
+ (B) x2a C
na+
l(C) x2a . n a C+l (D) none of these
B-2. The value of x55 5.5.5xx5
ò dx is equal to
(A) C)5n(
53
5x
+l
(B) C)5n(5 35x5
+l (C) C)5n(
53
5x5
+l
(D) none of these
B-3. The value of ò xcosxsinxtan
dx is equal to
(A) Cxtan2 + (B) Cxcot2 + (C) C2
xtan+ (D) none of these
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B-4. If ò- x
x
41
2dx = K sin–1 (2x) + C, then the value of K is equal to
(A) ln 2 (B) 2n21l (C)
21
(D) 2n1l
B-5. If y = ò + 2/32 )x1(dx
and y = 0 when x = 0, then value of y when x = 1, is
(A) 32
(B) 2 (C) 23 (D) 21
B-6. The value of ò tan3 2 x sec 2 x dx is equal to :
(A) 13 sec3 2 x –
12
sec 2 x + C (B) – 16
sec3 2 x – 12
sec 2 x + C
(C) 16 sec3 2 x –
12
sec 2 x + C (D) 13 sec3 2 x + 1
2sec 2 x + C
B-7. The value of ò + 2)xcosx(sinx2cos
dx is equal to :
(A) xcosxsin1
+-
+ C (B) ln (sin x + cos x) + C
(C) ln (sin x – cos x) + C (D) ln (sin x + cos x)2 + C
B-8. The value of ò a++ )]xtan(.xtan1[ dx is equal to :
(A) cos a . ln )xsin(xsin
a+ + C (B) tan a . ln )xsin(xsin
a+ + C
(C) cot a . ln xsec)xsec( a+
+ C (D) cot a . ln xcos)xcos( a+
+ C
B-9*. The value of mx nx2 . e dxò (when m, n Î N) is equal to :
(A) mx nx2 3 C
m n2 n n3+
++l l
(B) (m n2 n n3)xe
n n2 n n3
+
+
l l
l l+C (C) ( )
mx nx
m n
2 .3 Cn 2 ,3
+l
(D) x x(mn).2 .3 C
m n2 n n3+
+l l
Section (C) : Integration by Parts
C-1. The value of ò -- xe)1x( dx is equal to :
(A) –xex + C (B) xex + C (C) – xe–x + C (D) xe–x + C
C-2. The value of ò- xtan 1
e ÷÷ø
öççè
æ
+++
2
2
x1xx1
dx is equal to :
(A) x xtan 1e
- + C (B) x2 xtan 1
e-
+ C (C) x1
xtan 1e
- + C (D) none of these
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C-3. The value of ò ¢¢-¢¢ )]x(g)x(f)x(g)x(f[ dx is equal to :
(A) )x(g)x(f
¢ (B) f¢(x) g(x) – f(x) g¢(x)
(C) f(x) g¢(x) – f¢(x) g(x) (D) f(x) g¢(x) + f¢(x) g¢(x)
C-4*. If 3x 3xe cos 4xdx e (A sin4x Bcos4x) C= + +ò then :
(A) 4A = 3B (B) 2A = 3B (C) 3A = 4B (D) 4A + 3B = 1
Section (D) : Algebraic Integral
D-1. The value of ò ++ 1xxdx
2 is equal to :
(A) 23
tan–1 ÷÷ø
öççè
æ +
31x2
+ C (B) 32
tan–1 ÷÷ø
öççè
æ +
31x2
+ C
(C) 31
tan–1 ÷÷ø
öççè
æ +
31x2
+ C (D) none of these
D-2. The value of ò + 4/342 )1x(x1
dx is equal to
(A) Cx11
4/1
4 +÷ø
öçè
æ + (B) (x4 + 1)1/4 +C (C) Cx11
4/1
4 +÷ø
öçè
æ - (D) Cx11
4/1
4 +÷ø
öçè
æ +-
D-3. The value of 3
dxx 1 x-
ò is equal to
(A) 3
3
1 1 x 1n C3 1 x 1
- -+
- +l (B)
21 1 x 1n C3 1 x2 1
- ++
- -l (C) 3
1 1n C3 1 x
+-
l (D) 31 n 1 x C3
- +l
D-4. The value of ò +
-
1e1e
x
x
dx is equal to :
(A) ln ÷øöç
èæ -+ 1ee x2x – sec–1 (ex) + C (B) ln ÷
øöç
èæ -+ 1ee x2x + sec–1 (ex) + C
(C) ln ÷øöç
èæ -- 1ee x2x – sec–1 (ex) + C (D) none of these
D-5. If ò + 34 xxdx
= 2xA
+ xB
+ ln 1xx+ + C, then :
(A) A = 21
, B = 1 (B) A = 1, B = – 21
(C) A = – 21
, B = 1 (D) none of these
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Section (E) : Integration of Trigonometric Functions:
E-1. If ò xsinxcos
11
3
dx = – 2÷÷
ø
ö
çç
è
æ+
--
xtanBxtanA 25
29
+ C, then :
(A) A = 91
, B = 51-
(B) A = 91
, B = 51
(C) A = – 91
, B = 51
(D) none of these
E-2. The value of ò -1xsec dx is equal to :
(A) 2 ln ÷÷ø
öççè
æ-+
21
2xcos
2xcos 2
+ C (B) ln ÷÷ø
öççè
æ-+
21
2xcos
2xcos 2
+ C
(C) – 2 ln ÷÷ø
öççè
æ-+
21
2xcos
2xcos 2
+ C (D) none of these
E-3. The value of 3
dxcos x sin2xò is equal to
(A) 5 / 212 cosx tan x C
5æ ö+ +ç ÷è ø
(B) 5 / 212 tan x tan x C
5æ ö+ +ç ÷è ø
(C) 5 / 212 tan x tan x C
5æ ö- +ç ÷è ø
(D) none of these
Section (F) : Miscellaneous
F-1. If x x
x x
4e 6e dx9e 4e
-
-
+-ò = Ax + B nl |9e2x – 4| + C, then
(A) 3 35A ,F ,C 02 36
= - = = (B) 35 3A , B ,C R36 2
= = - Î
(C) 3 35A , B ,C R2 36
= - = Î (D) 3 35A , B ,C R2 36
= = Î
F-2*. Let f' (x) = 3x2 sin 1x
– x cos 1x
, if x ¹ 0 ; f(0) and f(1/p) = 0 then ;
(A) f(x) is continuous at x = 0 (B) f(x) is non derivable at x = 0(C) f' (x) is continuous at x = 0 (D) f' (x) is non derivable at x = 0
PART - II : SUBJECTIVE QUESTIONS
Section (A) : Integration Using Standard Integral
A-1. Integrate with respect to x:(i) (2x + 3)5 (ii) sin 2x (iii) sec2 (4x + 5) (iv) sec (3x + 2) (v) tan (2x + 1)
(vi) 23x + 4 (vii) 1
2x 1+(viii) e4x + 5 (ix) x 1+ (x)
12x 1+
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A-2. Integrate with respect to x :
(i) sin 2x + 1
x 1+(ii) tan (3x + 1) + e4x + 5 (iii) 2 tan (4x + 5) (iv) sin2 x
(v) cos2x (vi) sin 2x cos 3x (vii) 1
x 3 x 2+ - +
Section (B) : Integration Using Substitution
B-1. Integrate with respect to x :
(i) x sin x2 (ii) 2
xx 1+
(iii) sec2 x tan x (iv) x
x
e 1e x
++
(v) 1 sinxx cosx
-+
(vi) 2x
2x
ee 2-
(vii) 2
cos2x x 1x sin2x 2x
+ ++ +
(viii) sec x
n(sec x tanx)+l(ix)
xx 2+
(x) 2
xx
1ee
æ ö+ç ÷è ø(xi) (ex + 1)2 ex (xii) ( )5
1x x 1+
Section (C) : Integration by PartsC-1. Integrate with respect to x :
(i) x sin x (ii) x l n x (iii) x sin2x (iv) x tan–1 x (v) l nx (vi) sec3x
(vii) 23 x2x e (viii) 1sin x- (ix) 2 1
2
x tan x1 x
-
+(x) ex sin x (xi) ex (sec2x + tan x)
Section (D) : Algebraic IntegralD-1. Integrate with respect to x:
(i) 2x 4+ (ii) 2
1x 4+
(iii) 2
1x 4-
(iv) 2
1x 5+
(v) 2x 2x 5+ + (vi) 2
1x 2x 5+ +
(vii) 2(x 1) 1 x x- - - (viii) 2
2x 1x 3x 4
++ +
(ix) 5 3 3x a x+ (x) 15 5 5
1
x (1 x )+(xi)
2
4
x 8x
- (xii)3
3
x 1x x
-+
D-2. Integrate with respect to x :
(i) 1
(x 1)(x 2)+ + (ii) 2
1(x 1)(x 3)+ + (iii) 2
1(x 1) (x 2)+ + (iv)
1(x 1)(x 2)(x 3)+ + +
Section (E) : Integration of Trigonometric Functions:E-1. Integrate with respect to x :
(i) 1
2 cosx+(ii)
12 cosx-
(iii) 2sinx 2cosx3cos x 2sinx
++
(iv) 1
1 sinx cosx+ +
(v) 2
12 sin x+
(vi) 2cosec x.sinx
(sinx cosx)- (vii) 4
2
sin xcos x
Section (F) : MiscellaneousF-1. Integrate with respect to x :
(i) 4 2
1x x 1+ +
(ii) 2
4
1 x1 x
++
(iii) 2
2 4
1 x1 x x
-- +
F-2. Integrate with respect to x :
(i) 1
(x 1) x 2+ + (ii) 2
1(x 4) x 1- + (iii) 2
1(x 1) x 2+ + (iv) 2 2
1(x 1) x 2+ +
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PART - III : MISCELLANEOUS OBJECTIVE QUESTIONS
Match The column-I
1. Column – I Column – II
(A) If F(x) = x sinx1 cos x
++ò dx and F(0) = 0, (p)
2p
then the value of F(p/2) is :
(B) Let F(x) = 1sin x
2
xe 11 x
- æ ö-ç ÷ç ÷-è ø
ò dx and F(0) = 1, (q) 3p
If F(1/2) = p
p 6/e3k , then the value of k is :
(C) Let F(x) = 2 2dx
(x 1) (x 9)+ +ò and F(0) = 0, (r)4p
if F( 3 ) = 365
k, then the value of k is :
(D) Let F(x) = tanx
sin xcosxò dx and F(0) = 0 (s) p
if F(p/4) = pk2
, then the value of k is :
2. If I = dx
a b cosx+ò , where a, b > 0 and a + b = u, a – b = v, then match the following column.
Column – I Column – II
(A) v = 0 (p) I = uv1
ln
2xtanv–u
2xtanvu +
+ C
(B) v > 0 (q) I = uv2
tan–1 ÷÷ø
öççè
æ
2xtan
uv
+ C
(C) v < 0 (r) I = vu–
1 ln
2xtanv––u
2xtanv–u +
+ C
(s)u2 tan
2x + C
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Comprehension #1
Let In, m = n msin xcos x.dx.ò Then we can relate In, m with each of the following
(i) I n – 2 , m (ii) I n + 2 , m (iii) I n , m – 2 (iv) I n , m + 2
(v) I n–2 , m + 2 (vi) I n + 2 , m – 2
Suppose we want to establish a relation between In, m and In, m –2 , then we setP(x) = sinn + 1 x cosm – 1 x ..... (1)
In In, m and In , m – 2 the exponent of cos x is m and m – 2 respectively, the minimum of the two is m – 2, adding 1to the minimum we get m – 2 + 1 = m – 1 . Now choose the exponent m – 1 of cos x in P(x). Similarly choosethe exponent of sin x for P(x)Now differentiating both sides of (1), we getP' (x) = (n + 1) sinnx cosmx – (m – 1) sinn + 2 x cos m– 2 x
= (n + 1) sinnx cosmx – (m – 1) sinn x (1 – cos2x) cos m– 2 x= (n + 1) sinnx cosmx – (m – 1) sinn x cosm– 2 x (m –1) sinnx cosmx= (n + m) sinnx cosmx – (m – 1) sinn x cosm– 2 x
Now integrating both sides , we getsinn + 1 x cos m–1 x = (n + m) In, m – (m – 1) In , m – 2 .Similarly we can establish the other relations.
3. The relation between I4, 2 and I2, 2 is
(A) I4, 2 = 16 (– sin3x cos3x + 3I2, 2) (B) I4, 2 =
16 (sin3x cos3x + 3I2 , 2)
(C) I4 , 2 = 16 (sin3x cos3x – 3I2, 2) (D) I4 , 2 =
14
(– sin3x cos3x + 2I2 , 2)
4. The relation between I4, 2 and I6, 2 is
(A) I 4, 2 = 15 (sin5x cos3x + 8I6, 2) (B) I4, 2 =
15 (– sin5x cos3x + 8I6 , 2)
(C) I4, 2 = 15 (sin5x cos3x – 8I6, 2) (D) I 4, 2 =
16 (sin5x cos3x + 8I6 , 2)
5. The relation between I4, 2 and I4, 4 is
(A) I4, 2 = 13 (sin5 x cos3x + 8I4, 4) (B) I4, 2 =
13 (– sin5 x cos3x + 8I4, 4)
(C) I4, 2 = 13 (sin5 x cos3x – 8 I4, 4) (D) I4, 2 =
13 (sin5 x cos3x + 6 I4, 4)
Comprehension #2
It is known that
sinx cosx if 0 x2cos x sinxtanx cot x
sinx cosx 3if x2cosx sinx
ì p+ < <ï
ï+ = í- - pï + p < <ï - -î
( ) ( )d 1 3tanx cot x tanx cot x (tanx cot x), x 0, ,dx 2 2 2
p pæ ö æ ö- = + + " Î È pç ÷ ç ÷è ø è ø
and ( ) ( )d 1 3tanx cot x tanx cot x (tanx cot x), x 0, ,dx 2 2 2
p pæ ö æ ö+ = - + " Î È pç ÷ ç ÷è ø è ø
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6. Value of integral I = ( ) 3tanx cot x dx where x 0, ,2 2p pæ ö æ ö+ Î È pç ÷ ç ÷
è ø è øò is
(A) 1 tanx cot x2 tan C
2- æ ö-
+ç ÷ç ÷è ø
(B) 1 tanx cot x2 tan C
2- æ ö+
+ç ÷ç ÷è ø
(C) 1 tanx cot x2 tan C
2- æ ö-
- +ç ÷ç ÷è ø
(D) 1 tanx cot x2 tan C
2- æ ö+
- +ç ÷ç ÷è ø
7. Value of the integral I = ( )tanx cot x dx,where x 0,2pæ ö+ Î ç ÷
è øò , is
(A) 12 sin (cosx sinx) C- - + (B) 12 sin (sinx cosx) C- - +
(C) 12 sin (sinx cosx) C- - + (D) 12 sin (sinx cosx) C-- + +
8. Value of the integral I = ( )tan x cot x dx+ò , where x Î 3,2pæ öpç ÷
è ø is
(A) 2 sin–1 (cos x – sin x) + C (B) 2 sin–1 (sin x – cos x) + C
(C) 2 sin–1 (sin x + cos x) + C (D) – 2 sin–1 (sin x + cos x) + C
ASSERTION/REASONING
9. STATEMENT-1 : 5(sin x)ò cos x dx = 6
xsin6
+ C.
STATEMENT-2 : n(f(x)) f (x)¢ò = 1n
))x(f( 1n
+
+
+ C, n Î I.
(A) Statement -1 is true, Statement - 2 is true ; Statement - 2 is correct explanation for Statement - 1(B) Statement -1 is true, Statement - 2 is true ; Statement - 2 is NOT correct explanation for Statement - 1(C) Statement -1 is true, Statement - 2 is false.(D) Statement -1 is false, Statement - 2 is true.
10. STATEMENT-1 : If x > 0, x ¹ 1 then 2x x(log e (log e) ) dx-ò = x logxe + C.
STATEMENT-2 : xe (f(x) f (x)) dx+ ¢ò = ex f(x) + C and et = x iff t = lnx.
(A) Statement -1 is true, Statement - 2 is true ; Statement - 2 is correct explanation for Statement - 1(B) Statement -1 is true, Statement - 2 is true ; Statement - 2 is NOT correct explanation for Statement - 1(C) Statement -1 is true, Statement - 2 is false.(D) Statement -1 is false, Statement - 2 is true.
11. STATEMENT-1 : x
xn (e 1)
e+
òl
dx = x – ÷÷ø
öççè
æ +x
x
ee1
ln (ex + 1) + C.
STATEMENT-2 : f (x)f(x)¢
ò dx = ln |f(x)| + C.
(A) Statement -1 is true, Statement - 2 is true ; Statement - 2 is correct explanation for Statement - 1(B) Statement -1 is true, Statement - 2 is true ; Statement - 2 is NOT correct explanation for Statement - 1(C) Statement -1 is true, Statement - 2 is false.(D) Statement -1 is false, Statement - 2 is true.
Page No. # 12Arride learning Online E-learning AcademyA-479 Indra Vihar, Kota Rajasthan 324005
PART - I : OBJECTIVE QUESTIONS
1. If f(x) = ò-3x
x2sinxsin2dx, where x ¹ 0, then 0x
Limit® f'(x) has the value
(A) 0 (B) 1 (C) 2 (D) not defined
2. If ò -+
xtanxcot1x4cos
dx = A cos 4x + B; where A & B are constants, then
(A) A = – 1/4 & B may have any value (B) A = – 1/8 & B may have any value(C) A = – 1/2 & B = – 1/4 (D) none of these
3. The value of ( )xxx
e x+ò dx is equal to
(A) C]1xx[e2 x ++- (B) C]1x2x[e2 x ++-
(C) C]1xx[e2 x ++- (D) C]1xx[e2 x +++
4. The value of tane (sec sin )dq q - q qò is equal to
(A) tane sin Cq- q + (B) tane sin Cq q + (C) tane sec Cq q + (D) tane cos Cq q +
5. dx)x1(x
x17
7
ò +- equals :
(A) 72n| x | n|1 x | c7
+ + +l l (B) 72n| x | n|1 x | c4
- - +l l
(C) 72n| x | n |1 x | c7
- + +l l (D) 72n| x | n|1 x | c4
+ - +l l
6.1 cosx
cos cos x-a -ò dx where 0 < a < x < p , equals :
(A) c2xcos
2cosn2 +÷ø
öçèæ -
al (B) 1
xcos22 cos c
cos2
-
æ öç ÷
+ç ÷aç ÷ç ÷è ø
(C) c2xcos
2cosn22 +÷ø
öçèæ -
al (D) 1
xcos22sin c
cos2
-
æ öç ÷
- +ç ÷aç ÷ç ÷è ø
Page No. # 13Arride learning Online E-learning AcademyA-479 Indra Vihar, Kota Rajasthan 324005
7. The value of ò +- 4/153 ]2x()1x[(1
dx is equal to
(A) C2x1x
34 4/1
+÷ø
öçè
æ+-
(B) C1x2x
34 4/1
+÷ø
öçè
æ-+
(C) C2x1x
31 4/1
+÷ø
öçè
æ+-
(D) C1x1x
31 4/1
+÷ø
öçè
æ-+
8. The value of n sin x(x e cos x)-ò l dx is iqual to
(A) x cos x + C (B) sin x – x cos x + C(C) – e lnx cos x + c (D) sin x + x cos x + C
9. Antiderivative of 2
2
sin x1 sin x+
w.r.t. x is :
(A) ( )2x arctan 2 tanx C2
- + (B) 1 tanxx arctan C2 2
æ ö- +ç ÷
è ø
(C) ( )x 2 arctan 2 tanx C- + (D) tanxx 2 arctan C
2æ ö
- +ç ÷è ø
10. The value of dx2x3cos
2xcosxsin4ò is equal to
(A) Cx3cos31x2cos
21xcos ++- (B) Cx3cos
31x2cos
21xcos +--
(C) Cx3cos31x2cos
21xcos +++ (D) Cx3cos
31x2cos
21xcos +-+
11. The value of 1 x1 x
-
+ò dx is equal to
(A) ( )1x 1 x 2 1 x cos x C-- - - + + (B) ( )1x 1 x 2 1 x cos x C-- + - + +
(C) ( )1x 1 x 2 1 x cos x C-- - - - + (D) ( )1x 1 x 2 1 x cos x C-- + - - +
12. The value of ò )x16cos.x8cos.x4cos.x2cos.xcos.x(sin dx is equal to
(A) C1024
x16sin+ (B) C
1024x32cos
+- (C) C1096
x32cos+ (D) C
1096x32cos
+-
13. The value of 6 6
1cos x sin x+ò dx is equal to
(A) tan–1 (tan x + cos x) + C (B) – tan–1 (tan x + cos x) + C(C) tan–1 (tan x – cot x) + C (D) – tan–1 (tan x – cot x) + C
14. The value of xln(1 sinx) x tan
4 2ì üpæ ö+ + -í ýç ÷
è øî þò dx is equal to:
(A) x l n (1 + sin x) + C (B) l n (1 + sin x) + C(C) – x l n (1 + sin x) + C (D) l n (1 – sin x) + C
Page No. # 14Arride learning Online E-learning AcademyA-479 Indra Vihar, Kota Rajasthan 324005
15. The value of ò +-
2x1.
1x1x
dx is equal to
(A) Cx
1xx1sin
21 +
-+- (B) C
x1cos
x1x 1
2++
- -
(C) Cx
1xxsec2
1 +-
-- (D) Cx
1x1xtan2
21 +-
-+-
16. If I = 3sinx sin x dx
cos2x+
ò = A cos x + B l n |f(x)| + C, then
(A) 1 1 2 cos x 1A , B , f(x)4 2 2 cos x 1
- -= = =
+(B)
1 3 2 cos x 1A , B , f(x)2 4 2 2 cos x 1
- -= - = =
+
(C) 1 3 2 cos x 1A , B , f(x)2 2 2 cos x 1
+= - = =
-(D)
1 3 2 cos x 1A , B , f(x)2 4 2 2 cos x 1
- -= = =
+
17. If 3
3 5
dx a cot x b tan x Csin x cos x
= + +ò , where C is an arbitary constant of integration, then the
values of 'a' and 'b' are respectively:
(A) –2 & 23 (B) 2 & –
23 (C) 2 &
23 (D) none
More than one choice tyep
18. If 1dx xItan mtan C
5 4cosx 2- æ ö= +ç ÷+ è øò then :
(A) I = 2/3 (B) m = 1/3 (C) I = 1/3 (D) m = 2/3
19. The value of 2 2
2
x cos x1 x++ò cosec2x dx is equal to;
(A) cot x – cot–1 x + C (B) C – cot x + cot–1 x
(C) – tan–1 x – cosec x Csec x
+ (D) 1n tan xe cot x C-
- - +l
20. The value of 4 4sin2x dx
sin x cos x+ò is equal to :
(A) ( ) Cxcotcot 21 +- (B) ( ) Cxtancot 21 +- -
(C) ( ) Cxtantan 21 +- (D) ( ) Cx2costan 1 +- -
Page No. # 15Arride learning Online E-learning AcademyA-479 Indra Vihar, Kota Rajasthan 324005
21. The value of 2
dx
x x-ò is equals to
(A) cxsin2 1 +- (B) c)1x2(sin 1 +--
(C) )1x2(cos2c 1 -- - (D) cxx2cos 21 +--
22. The value of 2
x 1nx 1
x 1
-æ öç ÷+è ø
-òl
dx is equal to
(A) 21 x 1n C2 x 1
-+
+l (B) 21 x 1n C
4 x 1-
++
l (C) 21 x 1n C2 x 1
++
-l (D) 21 x 1n C
4 x 1+
+-
l
23. The value of n (tan x) dxsin xcosxòl is equal to
(A) 21 n (cot x) C2
+l (B) 21 n (sec x) C2
+l (C) 21 n (sinx sec x) C2
+l (D) 21 n (cos x cosec x) C2
+l
24. If In = ncot x dxò and I0 + I1 + 2 (I2 + .... + I8) + I9 + I10 = A 2 9u uu .......
2 9æ ö
+ + =ç ÷è ø
+ C, where u = cot x and
C is an arbitrary constant, then(A) A is constant (B) A = – 1 (C) A = 1 (D) A is dependent on X
PART - II : SUBJECTIVE QUESTIONS
Evaluate the following integrals:
1.1 dx
sin(x a)cos(x b)- -ò 2. tan x. tan2x. tan3x dxò 3. 3 3
x dxa x-ò
4.2 2
2 2
a xx dxa x
-+ò 5. ( )3 / 22
x nx dxx 1-
òl
6. 1 xsin dxa x
-
+ò
7. ò3
sin x2
xcos x sinxe .cos x
- dx 8. x x 1 dx
x 2+ +
+ò 9. 2
2sin2 cos6 cos 4sin
f - f- f - fò
10.2
6
4 x dxx+
ò 11. 4 4sin xcos x dxò 12. 4
1 dx1 sin x-ò
13.2
4 2
(x 1) dxx x 1
-+ +ò 14. ( )2 2sinx
1 xcosx dxx 1 x e
+
-ò
Page No. # 16Arride learning Online E-learning AcademyA-479 Indra Vihar, Kota Rajasthan 324005
PART-I IIT-JEE (PREVIOUS YEARS PROBLEMS)
1. Integrate, ( )
3
22
x 3x 2 dxx 1 (x 1)
+ +
+ +ò [IIT-JEE 1999], Part-2]
2. Let f(x) = xe (x 1)(x 2)- -ò dx then f decreases in the interval : [IIT-JEE 2000, ]
(A) (– ¥, 2) (B) (–2, –1) (C) (1, 2) (D) ),2( ¥+
3. Evaluate : 12
2x 2sin dx4x 8x 13
- æ ö+ç ÷ç ÷+ +è ø
ò . [IIT-JEE 2001]
4. For any natural number m, evaluate ( ) ( )1m3m 2m m 2m mx x x 2x 3x 6+ + + +ò dx , where x > 0.
[IIT-JEE 2002]
5.2
3 4 2
x 1 dxx 2x 2x 1
-
- +ò is equal to : [IIT-JEE 2006]
(A) 2 2 14 2
2x x
xc- +
+ (B) 2 2 14 2
3x x
xc- +
+ (C) 2 2 14 2x xx
c- ++ (D)
2 2 12
4 2
2x x
xc- +
+
6. Let f x x
xn nb ge j
=+1
1/ for n ³ 2 and gx f f f xf occurs n times
b g b g b g= ° ° °..... .1 244 344
Then ( )n 2x g x dx-ò equals :
[IIT-JEE 2007]
(A) 1
11
1 1
n nnx Kn n
-+ +
-
b g e j (B) 1
11
1 1
nnx Kn n
-+ +
-
b g e j
(C) 1
11
1 1
n nnx Kn n
-+ +
+
b g e j (D) 1
11
1 1
nnx Kn n
-+ +
+
b g e j
7. Let F(x) be an indefinite integral of sin .2 x [IIT-JEE 2007]Statement - 1 : The function F(x) satisfies F x F x+ =pb g b g for all real x.because
Statement - 2 : sin sin2 2x x+ =pb g for all real x.(A) Statement - 1 is True, Statement - 2 is True; Statement-2 is a correct explanation for Statement-1.(B) Statement - 1 is True, Statement - 2 is True; Statement - 2 is NOT a correct explanation for Statement - 1.(C) Statement - 1 is True, Statement - 2 is False.(D) Statement - 1 is False, Statement - 2 is True.
8. Let x
4x 2xeI dx
e e 1=
+ +ò , x
4x 2xeJ dx
e e 1
-
- -=+ +ò . Then for an arbitrary constant C, the value of J – I
equals : [IIT-JEE 2008]
(A) 4x 2x
4x 2x1 e e 1log2 e e 1
æ ö- +ç ÷
+ +è ø + C (B)
2x x
2x x1 e e 1log2 e – e 1
æ ö+ +ç ÷
+è ø + C
(C) 2x x
2x x1 e e 1log2 e e 1
æ ö- +ç ÷
+ +è ø + C (D)
4x 2x
4x 2x1 e e 1log2 e e 1
æ ö+ +ç ÷
- +è ø + C
Page No. # 17Arride learning Online E-learning AcademyA-479 Indra Vihar, Kota Rajasthan 324005
PART-II AIEEE (PREVIOUS YEARS PROBLEMS)
1. n
dxx(x 1)+ò is equal to : [AIEEE 2002]
(A) n1
log 1xxn
n
+ + c (B) n1
log n
n
x1x +
+ c (C) log 1xxn
n
+ + c (D) none of these
2. If ò a)–xsin(xsin
dx = Ax + B log sin (x – a) + c, then value of (A, B) is : [AIEEE 2004]
(A) (sin a, cos a) (B) (cos a, sin a) (C) (– sin a, cos a) (D) (– cos a, sin a)
3. ò xsin–xcosdx
is equal to : [AIEEE 2004]
(A) 21
log ÷øö
çèæ p
8–
2xtan + c (B)
21
log ÷øö
çèæ
2xcot + c
(C) 21
log ÷øö
çèæ p
83–
2xtan + c (D)
21
log ÷øö
çèæ p
+83
2xtan + c
4. ò ïþ
ïýü
ïî
ïíì
+
2
2)x(log1)1–x(log
dx is equal to : [AIEEE 2005]
(A) 1)x(log
x2 +
+ c (B) 2
x
x1xe+
+ c (C) 1x
x2 +
+ c (D) 1)x(logxlog2 +
5. ò + xsin3xcosdx
equals : [AIEEE 2007]
(A) 21
log tan ÷øö
çèæ p
+122
x + c (B)
21
log tan ÷øö
çèæ p
12–
2x
+ c
(C) log tan ÷øö
çèæ p
+122
x + c (D) log tan ÷
øö
çèæ p
12–
2x
+ c
6. The value of 2 ò÷øö
çèæ p
4–xsin
dxxsin is : [AIEEE 2008]
(A) x + log ÷ø
öçè
æ p4
–xcos + c (B) x – log ÷ø
öçè
æ p4
–xsin + x
(C) x + log ÷ø
öçè
æ p4
–xsin + c (D) x – log ÷ø
öçè
æ p4
–xcos + c
7. If the integral 5 tanx
sin x – 2cosxò dx = x + a ln | sinx – 2 cosx | + k, then a is equal to [AIEEE 2012]
(A) – 1 (B) – 2 (C) 1 (D) 2
Page No. # 18Arride learning Online E-learning AcademyA-479 Indra Vihar, Kota Rajasthan 324005
NCERT BOARD QUESTIONS
Verify the following:
1.2x 1 dx2x 3
-+ò = x – log |(2x + 3)2| +C 2. 2
2x 3 dxx 3x
++ò = log |x2 + 3x| + C
Evaluate the following:
3.2(x 2) dxx 1+
+ò 4.6logx 5logx
4logx 3logx
e e dxe e
--ò 5.
(1 cos x) dxx sinx++ò
6.dx dx
1 cosx+ò 7. 2 4tan x sec x dxò 8.sinx cosx dx
1 sin2x+
+ò
9. 1 sinx dx+ò 10.x dx
x 1+ò 11.a xa x
+-ò
12.
12
34
x dx1 x+
ò 13.2
4
1 x dxx+
ò 14. 2
dx dx16 9x-
ò
15. 2
dt
3t 2t-ò 16. 2
3x 1 dxx 9
-
+ò 17. 25 2x x dx- +ò
18. 4
x dxx 1-ò 19.
2
4
x dx1 x-ò put x2 = t 20. 22ax x dx-ò
21.1
32 2
sin x dx(1 x )
-
-ò 22.
(cos5x cos4x) dx1 2cos3x
+-ò 23.
6 6
2 2
sin x cos x dxsin xcos x
+ò
Page No. # 19Arride learning Online E-learning AcademyA-479 Indra Vihar, Kota Rajasthan 324005
EXERCISE # 1
PART # I
A-1. (A) A-2. (A) A-3. (B) A-4. (A) B-1. (B) B-2. (C) B-3. (A)B-4. (D) B-5. (D) B-6. (C) B-7. (B) B-8. (C) B-9*. (B , C) C-1. (C)C-2. (A) C-3. (C) C-4*. (C, D) D-1. (B) D-2. (D) D-3. (A) D-4. (A)D-5. (C) E-1. (B) E-2. (C) E-3. (B) F-1. (C) F-2*. (A, C, D)
PART # II
A-1. (i) 6(2x 3) C
12+
+ (ii) cos2x C
2- + (iii)
tan(4x 5) C4
++
(iv) 1 n | sec(3x 2) tan(3x 2) C |3
+ + + +l (v) 1 n | sec(2x 1) | C2
+ +l (vi) 3x 42 C
3 n2
+
+l
(vii) 1 n | 2x 1| C2
+ +l (viii) 4x 5e C4
+
+ (ix) 3/ 22(x 1) C
3+
+ (x) 2x 1 C+ +
A-2. (i) cos2x n | x 1| C
2- + + +l (ii) 4x 51 1n | sec(3x 1) | e C
3 4++ + +l
(iii) 1 n | sec(4x 5) | C2
+ +l (iv) x 1 sin2x C2 4
- +
(v) x 1 sin2x C2 4
+ + (vi) 1 1cos5x cos x C
10 2- + +
(vii) ( )3/ 2 3 / 22 (x 3) (x 2) C3
+ + + +
B-1. (i) 21 cos x C2
- + (ii) 21 n | x 1| C2
+ +l (iii) 21(tanx) C2
+ or 2sec x C
2+
(iv) xn | e x | C+ +l (v) n | x cos x | C+ +l (vi) 2x1 n | e 2 | C2
- +l (vii) 21 n | x sin2x 2x | C2
+ + +l
(viii) n | n(sec x tanx) | C+ +l l (ix) 3/ 2 1/ 22 (x 2) 4(x 2) C3
+ - + +
(x) 2x 2x1(e e ) 2x C2
-- + + (xi) 3x 2x x1 e e e C3
+ + +
(xii) 5
1 1n 1 C5 x
- + +l
Page No. # 20Arride learning Online E-learning AcademyA-479 Indra Vihar, Kota Rajasthan 324005
C-1 (i) sin x – x cos x + c (ii) 2 2x xn x C
2 4- +l (iii)
2x x 1sin2x cos2x C4 4 8
- - +
(iv) 2
1 1x x 1tan x tan x C4 2 2
- -- + + (v) x( nx 1) C- +l
(vi) sec x tanx 1 n | sec x tanx | C
2 2+ + +l (vii) 22 x(x 1) e C- +
(viii) 1 1x 1 x 1xsin x sin x C2 2
- --+ - + (ix)
1 21 21 (tan x)x tan x n (1 x ) C
2 2
-- - + - +l
(x) xe (sinx cosx) C
2- + (xi) ex tan x + C
D-1. (i) 2 2x x 4 2 n x x 4 C2
+ + + + +l (ii) 11 xtan C2 2
- +
(iii) 2n x x 4 C+ - +l (iv) 11 xtan C
5 5- +
(v) 2 2x 1 x 2x 5 2 n x 1 x 2x 5 C2+
+ + + + + + + +l (vi) 11 (x 1)tan C
2 2- +æ ö +ç ÷
è ø
(vii) 2 3 / 2
2 1(1 x x ) 3 15 2x 1(2x 1) 1 x x sin C3 8 8 5
- æ ö- - +- - + - - - +ç ÷
è ø
(viii) 2 14 2x 3n | x 3x 4 | tan C
7 7- +
+ + - +l (ix) 3
3 3 5 / 2 3 3 3 / 22 2a(a x ) (a x ) C15 9
+ - + +
(x) 4/ 5
5
1 11 C4 x
æ ö- + +ç ÷è ø(xi)
2 3 / 2
3
(x 8) C24x-
+
(xii) x – arctan x + 21 xn C
x+
+l
D-2. (i) x 1n Cx 2
++
+l (ii) 2 11 1 3n | x 3 | n | x 1| tan x C
10 20 10-+ - + + +l l
(iii) 1n | x 1| n | x 2 | C
(x 1)- + - + + +
+l l (iv)
1 1n | x 1| n | x 2 | n | x 3 | C2 2
+ - + + + + +l l l
E-1 (i) 12 tanx / 2tan C
3 3- æ ö
+ç ÷è ø
(ii) 12 xtan 3 tan C
23- æ ö +ç ÷
è ø
(iii) 10 2x n | 3cosx 2sinx | C13 13
- + +l (iv) xn 1 tan C2
+ +l
Page No. # 21Arride learning Online E-learning AcademyA-479 Indra Vihar, Kota Rajasthan 324005
(v) 11 3 tanxtan C
6 2- æ ö
+ç ÷ç ÷è ø
(vi) n |1 cot x | C- +l
(vii) 1 3xtanx sin2x C4 2
+ - +
F-1. (i) 2
1
1x 11 x 1 1 xtan n C142 3 3x x 1x
-+ -æ ö-
- +ç ÷è ø + +
l (ii) 2
11 x 1tan C2 2 x
- æ ö-+ç ÷ç ÷
è ø
(iii)
1x 31 xn C12 3 x 3x
+ -- +
+ +l
F-2. (i) x 2 1n Cx 2 1
+ -+
+ +l
(ii) 11 t 3 1n tan (t) C,24 3 t 3
--- +
+l where t = x 1+
(iii) 21 1 1 2n t t C,
3 3 93æ ö æ ö- - + - + +ç ÷ ç ÷è ø è ø
l where t = 1
x 1+
(iv) 2
12
x 2tan Cx
- +- +
PART # III
1. (A ® p), (B ® p), (C ® r), (D ® s)2. (A ® s), (B ® q), (C ® r)3. (A) 4. (A) 5. (B) 6. (A) 7. (B) 8. (A) 9. (C)10. (A) 11. (A)
EXERCISE # 2
PART # I
1. (B) 2. (B) 3. (C) 4. (D) 5. (C) 6. (D) 7. (A)8. (C) 9. (A) 10. (B) 11. (A) 12. (B) 13. (C) 14. (A)15. (C) 16. (D) 17. (A) 18. (A, B) 19. (B, C, D ) 20. (A , B, C, D)21. (A, B, D) 22. (B , D) 23. (A, C, D) 24. (A, B)
Page No. # 22Arride learning Online E-learning AcademyA-479 Indra Vihar, Kota Rajasthan 324005
PART # II
1.1 sin(x a)n C
cos(a b) cos(x b)-
+- -
l 2.1 1n | sec x | n | sec 2x | n | sec 3x | C2 3
é ù- - + +ê úë ûl l l
3.3 / 2
13 / 2
2 xsin C3 a
- æ ö+ç ÷
è ø4.
22 1 4 4
2
1 x 1a sin a x C2 2a
- æ ö+ - +ç ÷
è ø
5. arcsecx – 2
nx Cx 1
+-
l
6. (a + x) arc tanx ax Ca
- +
7. esinx (x–secx) + C 8. 1(x 1) 2 x 1 2 n | x 2 | 2 tan x 1 C-+ + + - + - + +l
9. 2 12 n sin 4sin 5 7 tan (sin 2) C-f - f + + f - +l 10.( ) ( )3 / 22 2
5
4 x x 6C
120x
+ -+
11.1 13x sin4x sin8x C
128 8é ù- + +ê úë û
12. ( )11 1tan 2 tanx tanx C22 2
- + +
13.2 2
1 11 x 1 2 2x 1tan tan C3 x 3 3 3
- -æ ö æ ö- +- +ç ÷ ç ÷
è ø è ø14. sin x 2 2sin x1n (xe ) n (1 x e ) C
2- - +l l
EXERCISE # 3
PART # I
1. 1 22
3 1 1 xtan x n (1 x) n(1 x ) C2 2 4 1 x
- - + + + + ++
l l
2. (C)
3. ( )1 22x 2 3(x 1) tan n 4x 8x 13 C3 4
- +æ ö+ - + + +ç ÷è ø
l
4.( )
m 13m 2m m m2x 3x 6x
C6 (m 1)
+
+ ++
+5. (D) 6. (A) 7. (D) 8. (C)
PART # II
1. (A) 2. (B) 3. (D) 4. (A) 5. (A) 6. (C) 7. (D)