200 bài tập tích phân - megabook.vn
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Trang 1
TP1: TÍCH PHÂN HÀM SỐ HỮU TỈ
xI dx
x x
2 2
21 7 12
Dạng 1: Tách phân thức
xxx ddd
x x
222 222
2
111 777 222
I dxx x
2
1
16 91
4 3
Câu 1. III xxxx x2 111
dddx x
222
1
x x x
2
116ln 4 9ln 3 III xxxx x
1
111666 999 111
444 333
xxx xxx 111111666 nnn 444 999lllnnn 333 1 25ln2 16ln3 = xxx
222lll 111 = 222555lllnnn222 111666lllnnn333
dxI
x x
2
5 31
.
x x
222
5 31
x
xx x x x3 2 3 2
1 1 1
( 1) 1
Câu 2. dddxxx
IIIx x5 3
1
xx x x x3 2 3 2
111 111
( 1) 1
I x xx
2
2
21 1 3 1 3ln ln( 1) ln2 ln5
2 2 2 812
Ta có: xxx
xx x x x3 2 3 2
111
( 1) 1
III xxx xxxx2
222111 111 333 111 333lllnnn lllnnn((( 111))) lllnnn222 lllnnn555
2 2 2 812
xI dx
x x x
5 2
3 24
3 1
2 5 6
x
222
2 2 2 2 812
xxxIII dddxxx
x x x
555 222
3 24
111
2 5 6
I
2 4 13 7 14ln ln ln2
3 3 15 6 5 Câu 3.
x x x3 24
333
2 5 6
444 111 777 111lllnnn lllnnn222
3 3 15 6 5
xdxI
x
1
0 3( 1)
III222 333 444
lllnnn3 3 15 6 5
ddd
x
111
0 3( 1)
x xx x
x x
2 3
3 3
1 1( 1) ( 1)
( 1) ( 1)
I x x dx
1 2 3
0
1( 1) ( 1)
8
Câu 4. xxx xxx
IIIx0 3( 1)
xxx xxx xxx xxx
x x
222 333
3 3((( 111))) ((( 111)))
( 1) ( 1)
ddd0
111 Ta có:
x x3 3
111 111
( 1) ( 1)
III xxx xxx xxx
111 222 333
0
111((( ))) ((( 111)))
888
xI dx
x
2
4
( 1)
(2 1)
Dạng 2: Đổi biến số
III dddx
222
4
((( 111
(2 1)
x xf x
x x
21 1 1
( ) . .3 2 1 2 1
Câu 5.
xxx xxx
x 4
)))
(2 1)
222111 111
3 2 1 2 1
xI C
x
31 1
9 2 1
Ta có:
xxx xxxfff xxx
xxx xxx
111((( ))) ... ...
3 2 1 2 1
xxxIII CCC
x
333111 111
9 2 1
xI dx
x
991
1010
7 1
2 1
x9 2 1
xxxIII dddxxx
x101
0
777 111
2 1
x dx x xI d
x x xx
99 991 1
20 0
7 1 1 7 1 7 1
2 1 9 2 1 2 12 1
Câu 6.
x
999999111
1010 2 1
dddxxx xxx xxxddd
x x xx
999
20 0
2 1 9 2 1 2 12 1
x
x
1001001 1 7 1 11
2 109 100 2 1 900
xI dx
x
1
2 20
5
( 4)
xxx III
x x xx
999999 999111 111
20 0
777 111 111 777 111 777 111
2 1 9 2 1 2 12 1
x
x
1001001 1 7 1 11
2 109 100 2 1 900
xxxIII dddxxx
x
111
2 20
555
( 4)
Câu 7.
x2 20 ( 4) t x2 4 I
1
8 Đặt ttt xxx
222444
111
888 III
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xI dx
x
1 7
2 50 (1 )
Trang 2
III dddx
111 777
2 5000 ((( t x dt xdx21 2 Câu 8.
xxx xxx
x2 5
111 )))dddttt222111
tI dt
t
2 3
5 51
1 ( 1) 1 1.
2 4 2
Đặt ttt xxx xxxdddxxx222 III dddttt
t
222 333
5 51
111 ((( 111))) 111 111
2 4 2
I x x dx1
5 3 6
0
(1 )
ttt
t5 51
...2 4 2
ddd111
0
(((
dt t tt x dt x dx dx I t t dt
x
1 7 83 2 6
20
1 1 11 3 (1 )
3 3 7 8 1683
Câu 9. III xxx xxx xxx555 333 666
0
111 )))
ttt ttt tttttt ddd
x
111 777 888333 222 666
20
111 111111 333 111 )))
3 3 7 8 1683
I dxx x
4 3
41
1
( 1)
Đặt ddd
ttt xxx dddttt xxx dddxxx dddxxx III ttt tttx2
0
111(((
3 3 7 8 1683
III dddxxxx x
444
41
111
( 1)
t x2Câu 10.
x x
333
41 ( 1)
222 tI dt
t t
3
21
1 1 1 3ln
2 4 21
Đặt ttt xxx
tttIII ddd
t t21
111lllnnn
2 4 21
dxI
x x
2
10 21 .( 1)
tttt t
333
21
111 111 333
2 4 21
xxxIII
x x10 2...((( 111)))
x dxI
x x
2 4
5 10 21
.
.( 1)
Câu 11.
ddd
x x
222
10 2111
ddd
x x
444
5 10 21 .( 1) t x5
xxx xxxIII
x x
222
5 10 21
...
.( 1)
555 dtI
t t
32
2 21
1
5 ( 1)
. Đặt ttt xxx
tttIII
t t
333222
2 21
111
555 ((( 111)))
xI dx
x x
2 7
71
1
(1 )
ddd
t t2 21
xxx ddd
x x
222 777
71 (1 )
x xI dx
x x
2 7 6
7 71
(1 ).
.(1 )
Câu 12. III xxx
x x71
111
(1 )
xxx xxx
x x
222 777 666
7 71
111
.(1 ) t x7 III dddxxx
x x7 71
((( )))...
.(1 )
777 tI dt
t t
128
1
1 1
7 (1 )
. Đặt ttt xxx
tttIII dddttt
t t
111 888
1
111 111
7 (1 )
dxI
x x
3
6 21 (1 )
t t
222
17 (1 )
dddxxxIII
x x
333
6 2111 )))
xt
1
Câu 13.x x6 2
111 (((
111 tI dt t t dt
t t
3
1634 2
2 21 3
3
11
1 1
Đặt : xxx
ttt III ddd dddttt
t t
333
111666333444 222
2 21 3
3
111111
1 1
117 41 3
135 12
ttt ttt ttt ttt
t t2 21 3
3
1 1
111 444
135 12
xI dx
x
2 2001
2 10021
.(1 )
= 111 777 111 333
135 12
xxx dddxxx
x
222 000000111
2 10021 (1 )
xI dx dx
x xx
x
2 22004
3 2 1002 10021 1 3
2
1. .
(1 ) 11
Câu 14. III x
222
2 10021
...(1 )
xxxIII ddd xxx
x x x
x
222 222444
3 2 1002 10021 1 3
2
... ...(1 ) 1
1
t dt dxx x2 3
1 21 xxx ddd
x x x
x
222000000
3 2 1002 10021 1 3
2
111
(1 ) 11
ttt dddttt dddxxxx x2 3
111 222 . Đặt
x x2 3111
x xdxI
x x
1 2000
2 2000 2 20
1 .2
2 (1 ) (1 )
.
xxx xxxdddxxxIII
x x
111 000
2 2000 2 20
111
222 (((111 ))) (((111 ))) Cách 2: Ta có:
x x
222000 000
2 2000 2 20
...222
t x dt xdx21 2
tI dt d
t tt t
10002 21000
1000 2 10011 1
1 ( 1) 1 1 1 11 1
2 2 2002.2
. Đặt ttt xxx dddttt xxxdddxxx222111 222
tttIII ddd
t tt t
000000
1000 2 10011 1
((( ))) 111 111 111111 111
2 2 2002.2
xI dx
x
2 2
41
1
1
ttt dddt tt t
111 000000222 222111 000000
1000 2 10011 1
111 111 111
2 2 2002.2
xxxIII dddxxx
x
222
41
111
1
Câu 15.
x
222
411
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x x
x xx
2 2
42
2
11
1
11
Trang 3
xxx xxx
x xx
222
42
2
111111
111
11
t x dt dxx x2
1 11
Ta có: x x
x
222
42
2
11
t x dt dx222
1 11
dtI dt
t tt
3 3
2 2
21 1
1 1 1
2 2 2 22
t
t
31 2 1 2 1
.ln ln22 2 2 2 2 2 11
. Đặt t x dt dxxxx xxx
1 11
dtI dt
3 3
2 2
111 111
1 1 1
t 3
1 2 1 2 1.ln ln2
222 222 111
xI dx
x
2 2
41
1
1
dt
I dtttt tttttt
3 3
2 2
222
1 1 1
222 222 222 222222
t
ttt
31 2 1 2 1
.ln ln2222 222 222 222111
xI dx
2 21
111
x x
x xx
2 2
42
2
11
1
11
Câu 16. x
I dxxxx
2 2
444111
1
xxx xxx
x xx
222
42
2
111111
111
11
t x dt dxx x2
1 11
Ta có:x x
x
222
42
2
11
t x dt dxxxx 222
1 11
dt
It
5
2
22 2
. Đặt t x dt dx xxx
1 11
dtI
5
2
2 2
dtI
ttt
5
2
222
2 2
dut u dt
u22 tan 2
cos
.
uuuttt uuu dddttt
u2222 aaannn 222
cos u u u u1 2
5 5tan 2 arctan2; tan arctan
2 2 Đặt
ddd
u2ttt
cos111 222aaarrr aaa ttt aaarrr
2 2
u
u
I d u u u2
1
2 1
2 2 2 5( ) arctan arctan2
2 2 2 2
; uuu uuu uuu uuu555 555
tttaaannn 222 cccttt nnn222;;; aaannn ccctttaaannn2 2
u
222
ddduuu uuu uuu
1
222 222 222 555aaarrrccc aaarrrccc
2 2 2 2
xI dx
x x
2 2
31
1
uuu
u
III
1
222 111((( ))) tttaaannn tttaaannn2222 2 2 2
dddx x3
1 xI dx
xx
2 2
1
11
1
Câu 17.
xxxIII xxx
x x
222 222
31
111
dddxxx
xx
222 222
11
t xx
1 Ta có: xxxIII
xx
1
111111
1
xxx
xxx
111. Đặt ttt I
4ln
5
xI dx
x
1 4
60
1
1
III 444
lllnnn555
xxxIII ddd
x
111
60
111
1
x x x x x x x x
x x x x x x x x
4 4 2 2 4 2 2 2
6 6 2 4 2 6 2 6
1 ( 1) 1 1
1 1 ( 1)( 1) 1 1 1
Câu 18. xxxx
444
60 1
xxx xxx xxx xxx xxx xxx xxx xxx
x x x x x x x x
444
6 6 2 4 2 6 2 6
111 ((( 111))) 111 111
1 1 ( 1)( 1) 1 1 1
d xI dx dx
x x
1 1 3
2 3 20 0
1 1 ( ) 1.
3 4 3 4 31 ( ) 1
Ta có:
x x x x x x x x
444 222 222 444 222 222 222
6 6 2 4 2 6 2 61 1 ( 1)( 1) 1 1 1
d xI dx dx
1 1 31 1 ( ) 1.
xI dx
x
3
23
40 1
d x
I dx dxxxx xxx
1 1 3
222 333 222000 000
1 1 ( ) 1.
333 444 333 444 333111 ((( ))) 111
xxxIII dddxxx
x
333
222333
4
000 111
xI dx dx
x x x x
3 3
23 3
2 2 2 20 0
1 1 1 1ln(2 3)
2 4 12( 1)( 1) 1 1
Câu 19. x4
xxx dddxxxx x x x
333 333
222333 333
2 2 2 20 0
111lllnnn(((222
2 4 12( 1)( 1) 1 1
xdxI
x x
1
4 20 1
xxx
III ddd x x x x2 2 2 2
0 0
111 111 111333)))
2 4 12( 1)( 1) 1 1
dddxxx
x x
111
4 2000
Câu 20. xxx
IIIx x4 2 111 t x2
dt dtI
t tt
1 1
2 220 0
1 1
2 2 6 31 1 3
2 2
. Đặt ttt xxx222ttt dddttt
IIIt t
t
111 111
2 220 0
111 111
2 2 6 31 1 3
2 2
ddd
t tt
2 220 0
2 2 6 31 1 3
2 2
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xI dx
x x
1 5
22
4 21
1
1
Trang 4
dddxxxx x
222
4 2
1 1
x x
x x xx
2 2
4 22
2
11
1
11 1
Câu 21. xxx
III x x
111 555
222
4 21
111
1
x x xx
222 222
4 22
2
111111
11 1
t x dt dxx x2
1 11
Ta có: xxx xxx
x x xx
4 22
2
111
11 1
t x dt dx222
1 11
dtI
t
1
20 1
. Đặt t x dt dxxxx xxx
1 11
dtI
1
222
0 1
dut u dt
u2tan
cos
dtI
ttt
1
0 1 ttt uuu ttt
u2ttt
cos I du
4
04
. Đặt
ddduuu ddd
u2aaannn
cos ddd
04
III uuu444
04
TP2: TÍCH PHÂN HÀM SỐ VÔ TỈ
xI dx
x x23 9 1
Dạng 1: Đổi biến số dạng 1
xxxIII ddd
x x23 9 1
xI dx x x x dx x dx x x dx
x x
2 2 2
2(3 9 1) 3 9 1
3 9 1
Câu 22. xxx
x x23 9 1
xxxIII xxx xxx xxx xxx xxx xxx xxx xxx xxx xxx
x x2(3 9 1) 3 9 1
3 9 1
I x dx x C2 31 13
ddd ddd ddd ddd
x x
222 222 222
2(3 9 1) 3 9 1
3 9 1
III xxx xxx111 111333 I x x dx22 9 1 x d x x C
32 2 2 2
2
1 19 1 (9 1) (9 1)
18 27 + xxx ddd CCC222 333
III xxx xxx xxx222 xxx ddd xxx xxx CCC111 111
999 111 (((999 ))) 999 111)))18 27
I x x C
32 32
1(9 1)
27
+ ddd222999 111
333222 222 222 222
222111 (((18 27
III xxx111
999 111)))27
x xI dx
x x
2
1
xxx CCC
333222 333222(((
27
xxxIII dddxxx
x x1
x xdx
x x
2
1
x xdx dx
x x x x
2
1 1
Câu 23. xxx
x x
222
1
xxx xxx
x x1
xxx xxxdddxxx xxx
x x x x1 1
xxx
dddx x
222
1
ddd
x x x x
222
1 1
xI dx
x x
2
11
.
xxxIII xxx
x x111
x x t x x21 1 x t3 2 2( 1) x dx t t dt2 24( 1)
3 + ddd
x x
222
111
xxx xxx ttt xxx xxx111 111 333 222 222 222 222
3
t dt t t C2 34 4 4( 1)
3 9 3
. Đặt t= 222 xxx ttt((( 111))) xxx dddxxx ttt ttt dddttt444
((( 111)))3
ddd ttt3 9 3 x x x x C
3
1
4 41 1
9 3 ttt ttt ttt CCC222 333444 444 444
((( 111)))3 9 3
xxx xxx xxx CCC333
1119 3
xI d x
x x2
1
= xxx 444 444
111 1119 3
xxxIII dddxxx
x x111
d x x
x x
2 (1 )
3 1
+
x x222
ddd
x x
222 (((111 )))
3 1
x x C2
41
3 =
xxx xxx
x x3 1 2223
I x x C3
41
9
= xxx xxx CCC444
1113
I x x C3
41
xI dx
x
4
0
2 1
1 2 1
Vậy: I x x C3
41
999
xxxIII dddxxx
x
444
01 2 1 t x2 1 Câu 24.
x0
222 111
1 2 1
ttt xxx
tdt
t
3 2
1
2 ln21
Đặt 222 111 ttt
dddtttt
333 222
1
222 lllnnn2221
. I =
t11
.
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dxI
x x
6
2 2 1 4 1
Trang 5
IIIx x
666
222 t x4 1 Câu 25.
dddxxx
x x
222 111 444 111 Đặt ttt xxx 444 111 I
3 1ln
2 12
I x x dx1
3 2
0
1
. III 333 111
lllnnn222 111222
ddd111
333 222
0
t x21 Câu 26. III xxx xxx xxx
0
111 ttt 111 I t t dt1
2 4
0
2
15 Đặt: xxx
222ttt ddd
111222 444
0
222
15III ttt ttt
015
xI dx
x
1
0
1
1
.
xxxIII dddxxx
x
111
0
111
1
t x
Câu 27.x01
xxx dx t dt2 . Đặt ttt ddd ddd...t t
dtt
1 3
0
21
xxx ttt ttt222
t
111 333
0
t t dtt
12
0
22 2
1
. I =
ttt tttdddttt
t0
222111
ddd
t
111222
0
114ln2
3= ttt ttt ttt
t0
222222 222
111
3=
111111444lllnnn222
3
xI dx
x x
3
0
3
3 1 3
.
dddx x
333
0 3 1 3
t x tdu dx1 2
Câu 28. xxx
III xxxx x0
333
3 1 3
ttt xxx tttddduuu xxx111 222 t t
I dt t dt dttt t
2 2 23
21 1 1
2 8 1(2 6) 6
13 2
33 6ln
2 Đặt ddd
ttt tttIII dddttt ttt dddttt dddttt
tt t
222 222 222
21 1 1
222 888 111((( 666 666
1113 2
333 333 666lllnnn
222
I x x dx0
3
1
. 1
tt t
333
21 1 1
222 )))3 2
ddd000
333
1
...
t tt x t x dx t dt I t dt
11 7 4
3 2 33
00
91 1 3 3( 1) 3
7 4 28
Câu 29. III xxx xxx xxx
1
111
ttt tttttt xxx ttt xxx dddxxx ttt dddttt III ttt ttt
00
999111 111 333 333((( 111))) 333
777 444 888
xI dx
x x
5 2
1
1
3 1
Đặt ddd
111111 777 444
333 222 333333
00222
xxxIII dddxxx
x x
555 222
1
111
3 1
tdtt x dx
23 1
3
Câu 30.x x1 3 1
tttxxx dddxxx
222 333 111
333
t
tdtI
tt
22
4
22
11
3 2.
31.
3
dtt dt
t
4 42
22 2
2( 1) 2
9 1
tt t
t
34 4
2 1 1 100 9ln ln .
9 3 1 27 52 2
Đặt tttddd
ttt
ttt
tttdddtttIII
t t
22
111111
333 222...
3331.
3
ddd
t
444 444222
22 2
222)))
999
tt t
t
34 4
2 1 1 100 9ln ln .
9 3 1 27 52 2
x xI dx
x
3 2
0
2 1
1
t
t
222222
444
22 1
.3
tttttt dddttt
t22 2
((( 111 222111
tt t
t
34 4
2 1 1 100 9ln ln .
9 3 1 27 52 2
x xI dx
xxx
3 2
000
2 1
111
x t x t21 1
Câu 31. x x
I dx3 22 1
222 dx tdt2 Đặt xxx ttt xxx ttt111 111 dddxxx tttdddttt222
t t tI tdt t t dt t
t
22 22 2 2 5
4 2 3
11 1
2( 1) ( 1) 1 4 542 2 (2 3 ) 2
5 5
tttIII tttddd ttt ttt dddttt ttt
t
222222 555
11 1
((( 111 555222 222 222 333 ))) 222
5 5
ttt ttt ttt
t
222 222222 222444 222 333
11 1
222 ))) ((( 111))) 111 444 444(((
5 5
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x dxI
x x
1 2
0
2( 1) 1
Trang 6
xxx dddxxxIII
x x
111
0
222( 1) 1
t x t x tdt dx21 1 2
t tI tdt t dt t
t tt
222 22 2 3
311 1
( 1) 1 1 16 11 2.2 2 2 2
3 3
Câu 32.x x
222
0 ( 1) 1
ttt ddd222
t tI tdt t dt t
t tt
222 22 2 3
311 1
( 1) 1 1 16 11 2.2 2 2 2
3 3
xI dx
x
4
20
1
1 1 2
Đặt ttt xxx ttt xxx dddttt xxx111 111 222
t tI tdt t dt t
t tt
222 22 2 3
311 1
( 1) 1 1 16 11 2.2 2 2 2
3 3
ddd
x2
0 1 1 2
dxt x dt dx t dt
x1 1 2 ( 1)
1 2
Câu 33.
xxxIII xxx
x
444
20
111
1 1 2
dddddd ttt
x111 ((( 111)))
1 2
t tx
2 2
2
Đặt
xxxttt xxx ttt dddxxx ttt ddd
x111 222
1 2
xxx
222 222
2
t t t t t tdt dt t dt
tt t t
4 4 42 3 2
2 2 22 2 2
1 ( 2 2)( 1) 1 3 4 2 1 4 23
2 2 2
và ttt ttt
2
ttt ttt ttt ttt ttt tttttt dddttt ttt dddttt
tt t t
222
2 2 22 2 2
111 ((( 222 222)))((( 111))) 111 333 444 222 111 444 222333
2 2 2
tt t
t
21 23 4ln
2 2
Ta có: I =
ddd tt t t
444 444 444333 222
2 2 22 2 2
2 2 2
ttt222
333 lllnnn2 2
= ttt
ttt ttt
111 222444
2 2
12ln2
4
xI dx
x
8
23
1
1
= 111
222lllnnn222444
III dddxxx
x2
3 1
xI dx
x x
8
2 23
1
1 1
Câu 34. xxx
x
888
23
111
1
x x2 23 1 1
x x x
82 2
31 ln 1
xxxIII dddxxx
x x
888
2 23
111
1 1
888
222 2223
1 ln 3 2 ln 8 3 = xxx xxx xxx 3
111 lllnnn 111
111 lllnnn 333 222 nnn 888 333
I x x x dx1
3 2
0
( 1) 2
= lll
III xxx xxx xxx dddxxx111
333 222
0
I x x x dx x x x x x dx1 1
3 2 2 2
0 0
( 1) 2 ( 2 1) 2 ( 1)
Câu 35.
0
((( 111))) 222
ddd ddd111 111
333 222 222 222
0 0
((( ))) 222 ((( 222 ))) 222 ((( ))) t x x22 III xxx xxx xxx xxx xxx xxx xxx xxx xxx xxx
0 0
111 111 111 ttt xxx xxx222 I2
15 . Đặt 222
15 III
222
15
x x xI dx
x x
2 3 2
20
2 3
1
.
xxx xxx xxxIII dddxxx
x x20
222 333
1
x x xI dx
x x
2 2
20
( )(2 1)
1
Câu 36.
x x
222 333 222
20 1
xxx xxx xxxIII xxx
x x
222
20
((( )))222 111)))
1
t x x2 1 I t dt3
2
1
42 ( 1)
3
ddd
x x
222
20
(((
1 ttt xxx xxx 111 III ttt dddttt
1
444222 ((( )))
333 . Đặt 222
333222
1
111
x dxI
x
2 3
3 20 4
.
ddd
x3 2
0 4
t x x t xdx t dt3 2 2 3 24 4 2 3
Câu 37. xxx xxx
III
x
222 333
3 20 4
ttt xxx xxx ttt dddxxx ttt ttt444 444 222 333 I t t dt3
24 3
4
3 3 8( 4 ) 4 2
2 2 5
Đặt xxx ddd
333 222 222 333 222 III ddd3 4
333(((
2 2 5
dxI
x x
1
211 1
ttt ttt ttt3
222444 333
4
333 888444 ))) 444 222
2 2 5
ddd
x x
111
211 1
Câu 38. xxx
III
x x211 1
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x x x xI dx dx
xx x
1 12 2
2 21 1
1 1 1 1
2(1 ) (1 )
xdx dx
x x
1 1 2
1 1
1 1 11
2 2
I dx x xx
11
1 11
1 1 11 ln | 1
2 2
xI dx
x
1 2
21
1
2
Trang 7
xxx xxx xxx xxx xxx xxx
222 222111 111
111 111 111 111
(((
xxxdddxxx xxx
111
111 111
111 111111
xxx111
222 222
xI dx
x
111 222
2221
1
2
t x t x tdt xdx2 2 21 1 2 2
Ta có: d dxx x
111 222
2 21 1
2(1 ) (1 ) ddd
x x
111 222
1 12 2
dddxxx
111111
111111
222 222
III xxx
1 2
2 222
tttdddttt ddd111 111 222
t dt
t
2 2
22
02( 1)
+ xxx xxx xxx111 111 111
nnn |||
xxx dddxxx
111
111
ttt xxx ttt xxx xxx xxx222 222 222 222 2
t dt2
0
I 1
+ tttdddttt ddd111 111 222ttt222
222((( )))
III
. Đặt ttt xxx ttt xxx xxx xxx222 222 222 222 dddttt
000222 111
111
t x x2 1
I2= ttt 222 222
t x x2 1
Vậy: 111
xxx xxx222 111
x xI dx
x
1
3 31
41
3
.
ttt xxx
xxxIII
xxx
1
3 31
I dxx x
11
3
2 31
3
1 11 .
Cách 2: Đặt
444111
333
III dddxxx
111111
333
222 333
333
111 111 t
x2
11
.
xxxdddxxx
111
111 ...
txxx
11 I 6Câu 39.
111
333 333111
III dddxxxxxx xxx
111111
333111 111 222
111111 III Ta có: 111 ...
xxx
xI dx
x
2 2
1
4
. Đặt ttt 666
xI dxxx
x
2
1
4
xI xdx
x
2 2
21
4
xxx
xxx
222
111
444
III xxxdddxxx
x t x tdt xdx2 2 24 4
.
III ddd
xxx
222 222
xxx ttt xxx dddttt xxxdddxxx444 444
t tdt t tdt dt t
tt t t
00 0 02
2 2 233 3 3
( ) 4 2(1 ) ln
24 4 4
Câu 40.
222
xxx x xxx
222111
444 ttt 222 222 222
ttt
000000 2
222 222 222333333 333
((( )))( ))) lll
222444
2 33 ln
2 3
Ta có: 222 222
xxx ttt xxx dddttt xxxdddxxx444 444
ttt
ttt
000 000
333
444 222
444 444
222 333333 lllnnn
222 333
xI dx
x x
2 5
2 22 ( 1) 5
. Đặt t = ttt222 222 222
tttdddttt ttt tttdddttt dddttt ttt
ttt ttt111 nnn
xxx
xxx x222 ((( )))
t x2 5
I =
222
(((222 333
333 lllnnn
ddd ttt xxx 555 dt
It
5
23
1 15ln
4 74
=
III xxx222 555
111
222 dtI
ttt
5
222333
1 15ln
444 777 Câu 41.
xI dx
xxx
2 5
222 222 555
ttt xxx 555 III444
xI dx
x x
27
3 21
2
Đặt 222 ddd555
111
xI dx
x x
2
3 21
2
ttt 111 555
lllnnn
xxx
7222
t x6t t
I dt dttt t t t
3 33
2 2 21 1
2 2 2 15 5 1
( 1) 1 1
2 55 3 1 ln
3 12
.
xxxIII dddxxx
222
333 222111
ttt ttt ttt
III dddttt tttt ttt222 222
111 111
222 222 222 111555 555 111
) 111 111
222lll
333 111
I dx
x x
1
20
1
1
Câu 42.
xxx
777
xxx666 dddtttttt ttt
333 333333
((( )))
555555 333 111 nnn
222
III
xxx000
111
t x x x2 1
Đặt ttt ttt ttt
III dddttt ttt ttt 222
222 222 222 111555 555 111
111
222lll
333 111
x x x dt
I tt
1 31 3
11
2 3 2 3ln(2 1) ln
2 1 3
ddd333 333333 555
555 333 111 nnn222
dddxxx
xxx
111
111
x x222 dddttt
III tttttt 1
111
222 333 222 333lllnnn(((222 111))) lllnnn
222
xI dx
x x
3 2
2 20 (1 1 ) (2 1 )
Câu 43. I dx
xxx
1
222
1
ttt xxx 111 111 333
111 333
111 333
xxx
222
222)))
Đặt xxx xxx222 dddtttIII ttt
111
222 333 222 333lllnnn(((222 111))) lllnnn
xxxIII xxx
111
x t2 1 I t dtt t
4
23
42 36 42 16 12 42ln
3
111 333
111 333
ddd xxx
333
222000 (((
x t2 1 t dtttt ttt
444
333
42 36 42 16 12 42llln
3
Câu 44. xxx
III xxx111 (((222 111 )))
III 222
Đặt xxx ttt222 111 ddd333
111 111 444 ttt ttt444222 666 444
222 666 222 222 nnn333
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xI dx
x x x x
3 2
0 2( 1) 2 1 1
Trang 8
dddx x x x
333 222
000 111 111
t x 1
Câu 45. xxx
III xxxx x x x
222((( ))) 222 111
t t dt
I t dtt t
2 22 22
21 1
2 ( 1)2 ( 1)
( 1)
t
23
1
2 2( 1)
3 3 Đặt ttt xxx 111
ddd ttt dddttt
t t
222 222222 222
21 1
)))111
( 1) ttt
1((( 111
3 3
x x xI d x
x
32 2 3
41
2011
ttt ttt ttt
III t t
222
21 1
222 ((( 111222 ((( )))
( 1)
222333
1
222 222)))
3 3
xxx xxx xxxIII dddxxx
x
222 222
41
222000 111
xI dx dx M Nx x
32 2 2 22
3 31 1
11
2011
xM dxx
32 2 2
31
11
Câu 46.
x
333 333
41
111
xxx ddd dddx x3 3
1 1
222000111
xM dxx
32 2 2
31
11
tx
32
11
Ta có: III xxx xxx MMM NNNx x
333222 222 222 222222
3 31 1
111111
111
xM dxx
32 2 2
31
11
t 32
11 M t dt
3 7
323
0
3 21 7
2 128
N dx x dxx x
2 22 2 2 23
3 21 1 1
2011 2011 140772011
162
. Đặt txxx
32
11 ddd
777
333222333
0
222
2 128
N dx x dxx x
2 22 2 2 23
3 21 1 1
2011 2011 140772011
162
I314077 21 7
16 128
MMM ttt ttt
333
0
333 111 777
2 128
N dx x dxx x
2 22 2 2 23
3 21 1 1
2011 2011 140772011
162
I314077 21 7
111 111222 I
314077 21 7
666 888
dxI
x x
1
33 30 (1 ). 1
.
xxxIII
x x33 3
0 (1 ). 1
t x3 31
Câu 47. ddd
x x
111
33 30 (1 ). 1
ttt 111 t dt
I dt
t t t t
3 32 22
2 21 14 3 2 33 3.( 1) .( 1)
dt dt tdt
t
tt ttt
3 3 3
2
3
2 2 2 3
2 2 41 1 1
3 342 333
11
111. 1
Đặt xxx333 333 dddttt
dddttt
t t t t
222 222222
2 21 14 3 2 33 3.( 1) .( 1)
dt dt t dt
t
tt t tt
3 3 3
2
3
2 2 2 3
2 2 41 1 1
3 342 333
11
111. 1
dtu du
t t3 4
1 31
ttt
III
t t t t
333 333
2 21 14 3 2 33 3.( 1) .( 1)
dt dt t dt
t
tt t tt
3 3 3
2
3
2 2 2 3
2 2 41 1 1
3 342 333
11
111. 1
dtu du
ttt ttt
1 31
u uI du u du u
111 12 1 2
2 1 22 23 33 3
30 0
00
1 1 1
13 3 3 23
Đặt dt
u du 333 444
1 31
uuu uuuIII ddduuu uuu uuu uuu
30 0
00
111 111 111
13 3 3 23
xI dx
x xx
2 2 4
231
1
ddd
111111111 111222 111 222
222 111 222222 222333 333333 333
30 0
00
13 3 3 23
xxxIII dddxxx
x xx
222 222
2
333111
111
Câu 48.
x xx
444
2
t x2 1 Đặt ttt xxx222
111
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tI dt
t
3 2 2
22
( 1)
2
t tdt t dt dt
t t
3 3 34 22
2 22 2 2
2 1 1 19 2 4 2ln
3 4 4 22 2
Trang 9
333 222 ttt
222 222222 222 222
999
x
I x x dxx
1
0
12 ln 1
1
ttt t 22
111
2
ttt ttt ttt ttt
ttt ttt
333 333 333
222 222222 222 222
111 222 444lllnnn
333 444 444 222222 222
xIII x x x
1
000
1222 ln 1
111
xH dx
x
1
0
1
1
= ddd 222
222111 111
xxx
HHH dxxxx
111
111x t tcos ; 0;
2
Dạng 2: Đổi biến số dạng 2
ddd
111
ooo 000;
H 22
Câu 49. xxx
xxx xxx xxx 111
111lllnnn 111
xxx dddxxx
000
xxx ttt tttccc sss ;;; ;
222
H 2
2
K x x dx1
0
2 ln(1 )
Tính 111
ooo 000
222
222
K x x dx1
0
2 ln(1 ) u x
dv xdx
ln(1 )
2
. Đặt xxx ttt ttt ccc sss ;;; ;;;222
HHH
111
000
222 lll ))) (((
K1
2
KKK xxx xxx dddxxxnnn(((111uuu xxxlllnnn 111 )))
K
1
2
I x x x dx2
5 2 2
2
( ) 4
Tính dddd
dd
vvvvvv
xxxxxx
(((
222
111
222
x x xdx2
5 2 2
2
)4
x x x dx2
5 2 2
2
( ) 4
. Đặt uuu xxx
dddxxx
lllnnn 111 )))
KKK
I x x x x222 222
222
( ) 4
x222
222
((( ))) 444
x x dx2
5 2
2
4
ddd222
555
xxx x dx222 2) 4 222
xxx xxx dx555 222
222
444
x x dx2
2 2
2
4
Câu 50. III xxx xxx xxx xxx ((( ))) 444
555 xxx xxx ddd222 dx2
x x dx2 2
222
4
I = xxx xxx 222
xxx xxx dddxxx555 222444222
dx222 222444
x x dx2
5 2
2
4
= xxx xxx dddxxx
2
x x x5 2
2
t x
+
d222
222
4
= A + B.
222
xxx xxx xxx555
ttt xxx
x x dx2
2 2
2
4
444 + Tính A = ddd
2
x x d x2 2
2
x t2sin
. Đặt ttt xxx
222222 222
222
xxx tttiiinnn
. Tính được: A = 0.
xxx xxx dddxxx 222sss 2+ Tính B = 444 xxx tttiiinnn 222
I 2
. Đặt 222sss
I 2
. Tính được: B =
222
x dxI
x
2 2
41
3 4
2
.
III
x xI
2 23 4
xI dx dx
x x
2 2 2
4 41 1
3 4
2 2
Vậy:
dddxxx444
111 222
xxx4 4
111 111
333 444
222 222
.
xxx
222 222 222
I1
Câu 51. xxx xxx
III 222 222333 444
III dddxxx dddxxx 444 444
111 dxx
2
41
3
2
Ta có: xxx
222 222 222
III xxx222
444 x dx2
4
1
3 7
2 16
.
dddxxx
333
222
2
x dx4
111
3 7 + Tính xxx222
111
222444
222 666
xI d x
x
2 2
2 41
4
2
= ddd 333
111
xI dx
x1
4
2
x t dx tdt2sin 2cos
= xxx dddxxx 333 777
666
xxx ddd
xxx111
444
222
xxx dddxxx ttt tttsssiiinnn 222ccc sss
.
III xxx ttt ddd222 ooo+ Tính 222 222
222 444xxx dddxxx ttt tttsssiiinnn 222ccc sss . Đặt ttt ddd 222 ooo .
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tdtI t dt t d t
t t
22 2 22 2
2 4 2
6 6 6
1 cos 1 1 1 3cot cot . (cot )
8 8 8 8sin sin
I1
7 2 316
Trang 10
tttddd ddd ttt
ttt
666 666 666
oooooo ccc
888 888 888 888iiinnn iiinnn
666
x dxI
x
1 2
60 4
ttt
III ttt ttt ttt ddd t t
222222 222 222222 222
222 444 222
6 6 6
111 ccc sss 111 111 111 333ccc ttt ooottt ... (((cccooottt )))
8 8 8 8sin sin
III 111666
x xI
1 2
t x dt x dx3 23
Vậy: 111
777 222 333
xxx666000 444
ttt xxxdt
I
t
1
20
1
3 4
.
ddd
dddxxx333 III
444
Câu 52. xxx xxx
III111 222
ttt xxx dddttt xxx dddxxx333 222 dddttt
ttt
111111
t u u dt udu2sin , 0; 2cos2
Đặt 333 III 222222000
333
t u u dt udu2sin , 0; 2c s2
I dt
6
0
1
3 18
dddttt111
111
222 ,,, 000 222 sss222
III ttt
.
ttt uuu uuu dddttt uuuddduuusssiii nnn ;;; ccc ddd666111
xI dx
x
2
0
2
2
Đặt ooo dddttt666
000
111
333 111888
x t dx tdt2cos 2sin
III
xxxIII
xxx
222
000
222
222
xxx ttt dddxxx ttt ttt222cccsss 222 iiinnn
tI dt
22
0
4 sin 22
.
dddxxx dddooo ssst
I dt2
24 sin 2
Câu 53. xxx ttt dddxxx ttt ttt222cccsss 222 iiinnn t
t
000
x dxI
x x
1 2
20 3 2
Đặt dddooo sss III dddsss222
x dxI
x x20 3 2
x dxI
x
1 2
2 20 2 ( 1)
ttt
ttt222
222444 iiinnn 222222
III
xxx222000 333 222
222 222000 222 ((( ) x t1 2cos
.
xxx dddxxx
xxx
xxx xxxIII
111 222
xxx ttt111 222ccc sss
Câu 54.
111 222
ddd
xxx 111
ooo
t tI dt
t
22
22
3
(1 2cos ) 2sin
4 (2cos )
Ta có: xxx xxx
III111 222
)))
xxx ttt111 222ccc sss
t tI dddt
t
22
22
3
(1 2s ) 2 in
4 (2cos )
t t dt
2
3
2
3 4cos 2cos2
. Đặt ooo
dddttt
ttt
222222
222222
333
((( 222 ))) 222 iiinnn
444 ((( )))
ttt dddttt
222
333
cccooosss
3 34
2 2
.
ttt tttIII
111 sss
222cccooosss ttt
222
333 444cccooosss 222 222
3 3
4
x x dx
1
22
0
1 2 1
dddcccooo sss
ttt dddttt
222
333
cccooosss
333 333
222 222
x dx
0
1 2 1 x tsin
= ttt 333 444cccooosss 222 222
444
000
111 222 111 xxx tttsssiiinnn I t t tdt6
0
3 1(cos sin )cos
12 8 8
=
xxx dddxxx I t t tdt6
000
3 1(cos sin )cos
12 8 8
Câu 55. xxx
111
222222 xxx tttsssiiinnn ttt )))s
222 888
I x dx3
2
2
1
Đặt tttdddtttooo nnn cccsss111
I d3
2
1
III ttt 666 333 111
(((ccc sss sssiii ooo888
333
222
111
Dạng 3: Tích phân từng phần
III dddCâu 56. xxx xxx 222
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xdu dxu x
xdv dxv x
221
1
xI x x x dx x dx
x x
3 32 2
2 22 2
3 11 . 5 2 1
2 1 1
dxx dx
x
3 32
22 2
5 2 1
1
I x x2 3
25 2 ln 1
Trang 11
xxxddd dddxxxxxx
dv dx v x
222111
xI x x x dx x dx
x x
3 32 2
2 22 2
3 11 . 5 2 1
2 1 1
dxx dx
x
3 32
22 2
5 2 1
1
I x x2 3
25 2 ln 1
I5 2 1
ln 2 1 ln22 4
Đặt uuu
uuu xxxdv dx
v x
222
111
xI x x x dx x dx
x x
3 32 2
2 22 2
3 11 . 5 2 1
2 1 1
dxx dx
x
3 32
22 2
5 2 1
1
I x x2 3
25 2 ln 1
I 5 2 1
ln 2 1 ln2222
xt
1
cos
I 5 2 1
ln 2 1 ln2444
tttcos 2;3 1;1 Chú ý: Không được dùng phép đổi biến xxx
111
cos;;;333 111;;; vì 222 111
http://megabook.vn/
Trang 12
TP3: TÍCH PHÂN HÀM SỐ LƯỢNG GIÁC
x xI dx
x x
28cos sin2 3
sin cos
Dạng 1: Biến đổi lượng giác
xxx xxxIII xxx
x x
222888cccsss sssiiinnn222 333
sin cos
x x x
I dx x x x x dxx x
2(sin cos ) 4cos2sin cos 4(sin cos
sin cos
x x C3cos 5sin
Câu 57. dddx x
ooo
sin cos
xxx xxxIII dddxxx xxx xxx xxx xxx xxx
x x
222iiinnn cccooosss ))) 444cccooo222sssiiinnn cccooosss sssiiinnn sss
sin cos
x x C3cos 5sin
xxx
dddx x
(((sss sss444((( cccooo
sin cos
x x C3cos 5sin
x x xI dx
x
cot tan 2tan2
sin4
.
dddx
ooo tttaaa tttaaa
sin4
x x x xI dx dx dx C
x x xx2
2cot 2 2tan2 2cot 4 cos4 12
sin4 sin4 2sin4sin 4
Câu 58. xxx xxx xxx
III xxxx
ccc ttt nnn 222 nnn222
sin4
xxxIII ddd ddd dddxxx
x x xx2
222cccooo tttaaa cccooo cccooosss222
sin4 sin4 2sin4sin 4
x
I dxx x
2cos8
sin2 cos2 2
Ta có: xxx xxx xxx
xxx xxx CCCx x xx2
ttt222 222 nnn222 222 ttt 444 444 111
sin4 sin4 2sin4sin 4
dddx x
222cccooo
sin2 cos2 2
x
I dx
x
1 cos 21 4
2 21 sin 2
4
xdx
dx
x x x
2
cos 21 4
2 21 sin 2 sin cos4 8 8
xdx
dx
x x2
cos 21 14
2 32 21 sin 2 sin
4 8
x x C1 3
ln 1 sin 2 cot4 84 2
Câu 59.
xxx
III xxxx x
sss888
sin2 cos2 2
xxx
III dddxxx
x
111 cccooosss 222111 444
2 21 sin 2
4
x dx
dx
x x x
2
cos 21 4
2 21 sin 2 sin cos4 8 8
x dx
dx
x x2
cos 21 14
2 32 21 sin 2 sin
4 8
x x C1 3
ln 1 sin 2 cot4 84 2
dxI
x x
3
2 3sin cos
Ta có:
x2 2
1 sin 24
x dx
dx
x x x
2
cos 21 4
2 21 sin 2 sin cos4 8 8
x dx
dx
x x2
cos 21 14
2 32 21 sin 2 sin
4 8
x x C1 3
ln 1 sin 2 cot4 84 2
dxI
xxx
333
dxI
x3
1
21 cos
3
Câu 60. dx
Ixxx 222 333sssiiinnn cccooosss
x3
111
21 cos
3
dx
Ix2
3
1
42sin
2 6
dddxxx
III
x3
2
1 cos3
xxxIII
2
3
111
42sin
2 6
1
4 3=
ddd
xxx2
3
42sin
2 6
111
4 3=
4 3
I dxx
6
0
1
2sin 3
.
dddx0 2sin 3
Câu 61. III xxx
x
666
0
111
2sin 3
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I dx dx
x x
6 6
0 0
11 1 22
sin sin sin sin3 3
x x
dx dxx x
x
6 6
0 0
coscos 2 6 2 63
sin sin 2cos .sin3 2 6 2 6
x x
dx dxx x
6 6
0 0
cos sin2 6 2 61 1
2 2sin cos
2 6 2 6
x x
6 60 0
ln sin ln cos .....2 6 2 6
Trang 13I
dx
dx
x
x6
6
0
0
1 1
1
22sin
sin
sin
sin 3
3
x
x
dx
dx
x
x
x6
6
0
0
cos cos
2 6
2 6
3 sin
sin
2cos
.sin 3
2 6
2 6
x
x
dx
dx
x
x6
6
0
0
cos
sin 2 6
2 6
1
1
2
2
sin
cos 2
6
2 6
x
x6
6
0
0
ln sin
ln cos
.
2 6
2 6
I x x x x dx2
4 4 6 6
0
(sin cos )(sin cos )
Ta có:
I dx dx
x x 6 6 0
0 1 1 1
222 sin
sin sin
sin 3 3
x x
dx dx x
x x 6 6
0 0 coscos 2
6 2
6 3sin sin
2cos
.sin 3 2
6 2
6
x x
dx
dx x
x 6
6 0
0 cos
sin 2 6
2 6
1 1 2 2
sin cos
2 6 2
6
x x6 6 0 0
ln sin
ln cos .
2 6
2 6
I x
x
x
x dx 2
4
4
6
6
000 (sin
cos )
(sin
cos )
Câu 62. I x x x x dx2
4 4 6 6(sin cos )(sin cos )
x x x x4 4 6 6(sin cos )(sin cos ) x x33 7 3
cos4 cos864 16 64
.
xxx xxx xxx xxx(((sssiiinnn cccooosss )))(((sssiiinnn cccooosss ))) 333
cccooosss sss64 16 64
I33
128 Ta có:
444 444 666 666 xxx xxx
333 777 333444 cccooo 888
64 16 64
333
111 888 III
333
222
I x x x dx2
4 4
0
cos2 (sin cos )
.
ddd
0
ccc (((sss ooo
I x x d x x d x2 2
2 2
0 0
1 1 1cos2 1 sin 2 1 sin 2 (sin2 ) 0
2 2 2
Câu 63. III xxx xxx xxx xxx222
444 444
0
ooosss222 iiinnn ccc sss )))
xxx xxx
0 0
ccc sss sss (((sss2 2 2
I x x dx2
3 2
0
(cos 1)cos .
III xxx xxx dddxxx ddd 222 222
222 222
0 0
111 111 111ooo 222 111 sssiiinnn 222 111 iiinnn 222 iiinnn222 ))) 000
2 2 2
ddd222
333 222
0
(((ccc ccc
x d x x d x2 2 2
5 2
0 0
cos 1 sin (sin )
Câu 64. III xxx xxx xxx
0
ooosss 111))) ooosss ...
dddxxx xxx ddd xxx
0 0
ooosss 111 iii nnn ))) 8
15 A = xxx
222 222 222555 222
0 0
ccc sssnnn (((sssiii
15
xdx x dx2 2
2
0 0
1cos . (1 cos2 ).
2
= 888
15
xdx x dx2 2
2 1cos . (1 cos2 ).
222
4
B = xdx x dx
2 22
000 000
1cos . (1 cos2 ).
4
8
15
= 4
15 4
Vậy I =
888
15 444
–
22
0
I cos cos 2x xdx
.
222 222
0
III cccooo cccoooxxx xxxddd
I x xdx x xdx x x dx2 2 2
2
0 0 0
1 1cos cos2 (1 cos2 )cos2 (1 2cos2 cos4 )
2 4
Câu 65.
0
sss sss 222 xxx
xxx xxx xxx xxxdddxxx xxx xxx ddd222 222
222
0 0 0
ooosss cccooo222 (((111 cccooo222 sss 111 cccooo222 cccooo 4442 4
III ddd xxx xxx222
0 0 0
111 111ccc sss sss )))cccooo 222 ((( 222 sss sss )))
2 4
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x x x2
0
1 1( sin2 sin4 )
4 4 8
xI dx
x
32
0
4sin
1 cos
Trang 14
xxx 222
000
((( sssiiinnn iii888
xxx dddxxx
x
333
0
444sssiiinnn
1 cos
x x xx x x x x
x x
3 3
2
4sin 4sin (1 cos )4sin 4sin cos 4sin 2sin2
1 cos sin
I x x dx20
(4sin 2sin2 ) 2
Câu 66. III x
2220 1 cos
x x2
sss 444444sss 444 ccc sss
1 cos sin
I x x dx20
(4sin 2sin2 ) 2
I xdx2
0
1 sin
xxx xxx xxx
xxx xxx xxx xxx xxxx x
333 333
2
444 iiinnn sssiiinnn (((111 cccooosss )))iiinnn sssiiinnn ooosss 444sssiiinnn 222 iiinnn222
1 cos sin
I x x dx20
(4sin 2sin2 ) 2
III xxxdddxxx222
0
1 sin
x x x xI dx dx
22 2
0 0
sin cos sin cos2 2 2 2
x
dx2
0
2 sin2 4
x xdx dx
3
22
30
2
2 sin sin2 4 2 4
4 2
Câu 67.
0
1 sin
III xxx ddd
222222 222
0 0
sssiii ccc sss2 2 2 2
ddd222
0
sss
x xdx dx
3
22
30
2
2 sin sin2 4 2 4
4 2
dxI
x
4
60 cos
xxx xxx xxx xxx
ddd xxx
0 0
nnn ooosss iiinnn cccooosss2 2 2 2
xxx xxx
0
222 iiinnn222 444
x xdx dx
3
22
30
2
2 sin sin2 4 2 4
4 2
x60 cos
I x x d x4
2 4
0
28(1 2tan tan ) (tan )
15
Câu 68. dddxxx
IIIx
444
60 cos
444
222 444
0
aaa aaa aaa111
Ta có: III xxx xxx ddd xxx
0
222888(((111 222ttt nnn ttt nnn ))) (((ttt nnn )))
555
.
xdxI
x x
sin2
3 4sin cos2
Dạng 2: Đổi biến số dạng 1
xxxddd
3 4sin cos2
x xI dx
x x2
2sin cos
2sin 4sin 2
Câu 69. xxx
III xxx xxx
sssiiinnn222
3 4sin cos2
xxx xxxIII xxx
2 x x
iiinnn ccc sss
2sin 4sin 2
t xsin Ta có: ddd
2 x x
222sss ooo
2sin 4sin 2 ttt xxxiiinnn I x C
x
1ln sin 1
sin 1
. Đặt sss III xxx CCC
x
111 nnn sssiiinnn 111
sssiiinnn 111
dxI
x x3 5sin .cos
x
lll
dddxxxIII
3 x x5sin .cos
xx
dx
xxx
dxI
23233 cos.2sin8
cos.cos.sin
Câu 70. 3 x x5sin .cos
xxxx 222333222333333 ccc.2sincos.cos.sin
t xtan
xxxx
dddxxx
xxx
dddxxxIII
ooosss.2sin888
cos.cos.sin
tttaaa I t t t dt x x x Ct x
3 3 4 2
2
3 1 3 13 tan tan 3ln tan
4 2 2tan
Đặt ttt xxxnnn III ttt ttt ttt ttt xxx xxx xxx CCC
t x
333 333 444 222
2
333 111 333 111333 tttaaannn aaannn 333lllnnn ttt nnn
4 2 2tan
tx
t2
2sin2
1
. ddd t x2
ttt aaa4 2 2tan
tttxxx
t2
222iiinnn222
1
Chú ý:
t2sss
1.
http://megabook.vn/
dxI
x x3sin .cos
Trang 15
dddxxxIII
x x3sin .cos
dx dxI
x x x x x2 22
sin .cos .cos sin2 .cos
Câu 71.x x3sin .cos
xxx xxxIII
x x x x x2 2222
sin .cos .cos sin2 .cos t xtan
dx tdt x
x t2 2
2; sin2
cos 1
dt tI dt
t t
t
2
2
12
2
1
t x
t dt t C x Ct
2 21 tan( ) ln ln tan
2 2
ddd ddd
x x x x x2 2sin .cos .cos sin2 .costttaaa
dddxxx tttdddttt xxx
2 x t2222
;;; sssiiinnn222cos 1
dt tI dt
t t
t
2
2
12
2
1
t x
t dt t C x Ct
2 21 tan( ) ln ln tan
2 2
x xI xdx
x
2011 2011 2009
5
sin sincot
sin
. Đặt ttt xxxnnn 2 x t2cos 1
dt tI dt
t t
t
2
2
12
2
1
t x
t dt t C x Ct
2 21 tan( ) ln ln tan
2 2
xxx xxxIII dddxxx
x
111 000111 000000
5
iiinnn iiinnnccc ttt
sin
xxI xdx xdxx x
2011 2011 22
4 4
11
cotsin cot cotsin sin
Câu 72. xxxx
222000 111 222 111 222 999
5
sss sssooo
sin
III xxxdddxxx xxxdddx x
111 222000111111 222222
4 4
111tttiii ttt
sin sin
t xcot
Ta có: xxxxxx xxx
x x
222000111
4 4
111
cccooosss nnn cccooo cccoootttsin sin
ccc I t tdt t t C
2 4024 804622011 2011 2011
2011 2011t (1 )
4024 8046 Đặt ttt xxxooottt III tttddd
444000222 888000444
000111 222000111 222000111000111 222000111
ttt (((000222 888000444
x x C
4024 8046
2011 20112011 2011
cot cot4024 8046
ttt ttt ttt ttt CCC
222 444 666222222 111 111 111
222 111 111 111 )))
444 444 666
xxx xxx CCC
444000222 888000444
000111 222000111000111 222000111
cccooottt cccooo4024 8046
x xI dx
x
2
0
sin2 .cos
1 cos
=
444 666
222 111 111222 111 111
ttt4024 8046
xxx xxxIII xxx
x
222
0
sssiiinnn ...cccooosss
1 cos
x xI dx
x
22
0
sin .cos2
1 cos
Câu 73. dddx
0
222
1 cos
x0
sss
ccct x1 cos Ta có:
xxx xxxIII dddxxx
x
222222
0
iiinnn ...cccooosss222
111 ooosss
ttt xxx111 ooosss t
I dtt
2 2
1
( 1)2 2ln2 1
. Đặt ccc
tttIII dddttt
t
222 222
1
((( 111)))222 222lllnnn222 111
I x xdx3
2
0
sin tan
t
1
xxxddd
0
tttaaa
x x xI x dx dx
x x
23 32
0 0
sin (1 cos )sinsin .
cos cos
Câu 74. III xxx xxx333
222
0
sssiiinnn nnn
xxx xxxxxx dddxxx dddxxx
x x
222
0 0
111 ooosss )))sssiiinnnnnn ...
cccsss ccc
t xcos Ta có:
xxx III
x x
222333 333
0 0
sssiiinnn ((( ccc sssiii
ooo ooosss xxxccc
uI du
u
1
22
1
1 3ln2
8
. Đặt ttt ooosss
dddu
222
1
I x x dx2
2
sin (2 1 cos2 )
uuu
III uuuu
111
222
1
111 333lllnnn222
888
ddd222
2
sin (2 1 cos2 )
I xdx x xdx H K2 2
2 2
2sin sin 1 cos2
Câu 75. III xxx xxx xxx
2
sin (2 1 cos2 )
222 xxxdddxxx xxx xxxdddxxx HHH KKK222
2 2
2sin sin 1 cos2
Ta có: III
2 2
2sin sin 1 cos2
http://megabook.vn/
H xdx x dx2
2 2
2sin (1 cos2 )2 2
Trang 16
222 xxxddd ddd
2 2
2sin (1 cos2 )2 2
K x x x xdx2 2 2
2 2
sin 2cos 2 sin cos
xd x2
2
22 sin (sin )
3
I2
2 3
+ HHH xxx xxx xxx
2 2
2sin (1 cos2 )2 2
xxxddd222 222 222
2 2
sin 2cos 2 sin cos
222 xxxddd xxx
2
sssnnn (((sssiiinnn )))
I 2
2 3
dxI
x x
3
2 4
4
sin .cos
+ KKK xxx xxx xxx xxx
2 2
sin 2cos 2 sin cos
2
222222 iii
333
I 2
2 3
dxI
xxx xxx
3
sssiiinnn ...ccc sss
dxI
x x
3
2 2
4
4.sin 2 .cos
Câu 76. dx
I3
222 444
444
ooo
dddxxxIII
2 x x2
4
...sin 2 .cos
t xtanx x
333
2 2
4
444sin 2 .cos
tttaaa
dxdt
x2cos. Đặt ttt xxxnnn dddttt
x2cos
t dt tI t dt t
tt t
33 32 2 3
2
2 211 1
(1 ) 1 1 8 3 42 2
3 3
dddxxx
x2cos
t dt tI t dt t
tt t
33 32 2 3
2
2 211 1
(1 ) 1 1 8 3 42 2
3 3
2
2
0
sin 2
2 sin
xI dx
x
.
t dt tI t dt t
tt t
33 32 2 3
2
2 211 1
(1 ) 1 1 8 3 42 2
3 3
2
2
sin 2
sssiii
xI dx
x x xI dx dx
x x
2 2
2 20 0
sin2 sin cos2
(2 sin ) (2 sin )
Câu 77.
2
2
000
sin 2
222 nnn
xI dx
xxx
ddd dddx x2 2
0 0
sss
(2 sin ) (2 sin ) t x2 sin Ta có:
xxx xxx xxxIII xxx xxx
x x
222 222
2 20 0
sssiiinnn222 iiinnn cccooosss222
(2 sin ) (2 sin )
ttt xxx222 sssiiinnn . Đặt
tI dt dt t
t tt t
33 3
2 22 2 2
2 1 2 22 2 2 ln
3 22ln
2 3
.
ttt ddd ddd ttt
t tt t
333 333
2 22 2 2
222 222222 222 222 lll
333 222
222
xI dx
x
6
0
sin
cos2
III ttt ttt t tt t
333
2 22 2 2
111 222nnn
222lllnnn
333
dddx
0cos2
x xI dx dx
x x
6 6
20 0
sin sin
cos2 2cos 1
Câu 78. xxx
III xxxx
666
0
sssiiinnn
cos2
xxx xxxIII dddxxx dddxxx
x x20 0
sss nnn sss nnn
cos2 2cos 1
t x dt xdxcos sin x x
666 666
20 0
iii iii
cos2 2cos 1 ttt xxx dddttt xxx xxxcccooosss sssiiinnn
x t x t3
0 1;6 2
. Đặt ddd
xxx ttt xxx ttt 333
000 111;;;666 222
tI dt
tt
31
2
231
2
1 1 2 2ln
2 2 2 22 1
Đổi cận:
tttddd
tt
333111
222
231
2
111 111 222lll
2 2 2 22 1
1 3 2 2ln
2 2 5 2 6
Ta được III ttt
tt231
2
222nnn
2 2 2 22 1
333 222
2 2 5 2 6
=
111 222lllnnn
2 2 5 2 6
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Trang 17
xI e x x dx22
sin 3
0
.sin .cos .
Câu 79.xI e x x dx
22sin 3
0
.sin .cos .
t x2sin Đặt t x2sin te t dt1
0
1(1 )
2 e
11
2= e
11
2 .
I x x dx2 12sin sin
2
6
Câu 80. I x x dx2 12sin sin
2
6
t xcos Đặt t xcos I3
( 2)16
. I 3
( 2)16
xI dx
x x
4
6 60
sin4
sin cos
Câu 81.
xI dx
x x
4
6 60
sin4
sin cos
xI dx
x
4
20
sin4
31 sin 2
4
x
I dx
x
4
20
sin4
31 sin 2
4
t x231 sin 2
4 . Đặt t x23
1 sin 24
dtt
1
4
1
2 1
3
I = dt
t
1
4
1
2 1
3
t
1
1
4
4 2
3 3= t
1
1
4
4 2
3 3 .
xI dx
x x
2
30
sin
sin 3cos
Câu 82.
xI dx
x x
2
30
sin
sin 3cos
x x xsin 3cos 2cos6
Ta có: x x xsin 3cos 2cos
6
x xsin sin6 6
;
x xsin sin6 6
x x
3 1sin cos
2 6 2 6
= x x
3 1sin cos
2 6 2 6
x dxdx
x x
2 2
3 20 0
sin63 1
16 16cos cos
6 6
I =
x dx dx
x x
2 2
3 20 0
sin63 1
16 16cos cos
6 6
3
6=
3
6
x xI dx
x
24
2
3
sin 1 cos
cos
Câu 83.
x xI dx
x
24
2
3
sin 1 cos
cos
x xI x dx x dx
x x
4 42
2 2
3 3
sin sin1 cos . sin
cos cos
x x
x dx x dxx x
0 4
2 20
3
sin sinsin sin
cos cos
x x
I x dx x dxx x
4 42
2 2
3 3
sin sin1 cos . sin
cos cos
x x
x dx x dxx x
0 4
2 20
3
sin sinsin sin
cos cos
x xdx dx
x x
0 2 24
2 20
3
sin sin
cos cos
7
3 112
=
x xdx dx
x x
0 2 24
2 20
3
sin sin
cos cos
7
3 112
.
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Trang 18
I dxx x
6
0
1
sin 3cos
Câu 84. I dxx x
6
0
1
sin 3cos
I dxx x
6
0
1
sin 3cos
I dxx x
6
0
1
sin 3cos
dx
x
6
0
1 1
2sin
3
= dx
x
6
0
1 1
2sin
3
x
dx
x
6
20
sin1 3
21 cos
3
=
x
dx
x
6
20
sin1 3
21 cos
3
.
t x dt x dxcos sin3 3
Đặt t x dt x dxcos sin
3 3
I dt
t
1
2
20
1 1 1ln3
2 41
I dt
t
1
2
20
1 1 1ln3
2 41
I x xdx2
2
0
1 3sin2 2cos
Câu 85. I x xdx2
2
0
1 3sin2 2cos
I x x dx2
0
sin 3cos
I x x dx2
0
sin 3cos
I x x dx x x dx3 2
0
3
sin 3cos sin 3cos
3 3 = I x x dx x x dx3 2
0
3
sin 3cos sin 3cos
3 3
xdxI
x x
2
30
sin
(sin cos )
Câu 86.xdx
Ix x
2
30
sin
(sin cos )
x t dx dt2
Đặt x t dx dt
2
tdt xdxI
t t x x
2 2
3 30 0
cos cos
(sin cos ) (sin cos )
tdt xdx
It t x x
2 2
3 30 0
cos cos
(sin cos ) (sin cos )
dx dx2I x
x x x
2 2 4
22 00 0
1 1cot( ) 1
2 2 4(sin cos ) sin ( )4
dx dx2I x
x x x
2 2 4
22 00 0
1 1cot( ) 1
2 2 4(sin cos ) sin ( )4
I
1
2 I
1
2
x xI dx
x x
2
30
7sin 5cos
(sin cos )
Câu 87.
x xI dx
x x
2
30
7sin 5cos
(sin cos )
xdx xdxI I
x x x x
2 2
1 23 30 0
sin cos;
sin cos sin cos
Xét:
xdx xdxI I
x x x x
2 2
1 23 30 0
sin cos;
sin cos sin cos
.
x t2
Đặt x t
2
. Ta chứng minh được I1 = I2
dx dxx
x x x
2 2
220 0
1tan( ) 12
2 4sin cos 02cos ( )4
Tính I1 + I2 =
dx dx x
x x x
2 2
220 0
1tan( ) 12
2 4sin cos 02cos ( )4
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Trang 19
I I1 2
1
2 I I I1 27 –5 1 I I1 2
1
2 I I I1 27 –5 1 .
x xI dx
x x
2
30
3sin 2cos
(sin cos )
Câu 88.
x xI dx
x x
2
30
3sin 2cos
(sin cos )
x t dx dt2
Đặt x t dx dt
2
t t x xI dt dx
t t x x
2 2
3 30 0
3cos 2sin 3cos 2sin
(cos sin ) (cos sin )
t t x xI dt dx
t t x x
2 2
3 30 0
3cos 2sin 3cos 2sin
(cos sin ) (cos sin )
x x x xI I I dx dx dx
x x x x x x
2 2 2
3 3 20 0 0
3sin 2cos 3cos 2sin 12 1
(sin cos ) (cos sin ) (sin cos )
x x x xI I I dx dx dx
x x x x x x
2 2 2
3 3 20 0 0
3sin 2cos 3cos 2sin 12 1
(sin cos ) (cos sin ) (sin cos )
I
1
2 I
1
2 .
x xI dx
x20
sin
1 cos
Câu 89.x x
I dxx2
0
sin
1 cos
t t tx t dx dt I dt dt I
t t2 20 0
( )sin sin
1 cos 1 cos
t d tI dt I
t t
2
2 20 0
sin (cos )2
4 4 81 cos 1 cos
Đặt t t t
x t dx dt I dt dt I2 t t2
0 0
( )sin sin
1 cos 1 cos
t d tI dt I
t t
2
2 20 0
sin (cos )2
4 4 81 cos 1 cos
x xI dx
x x
42
3 30
cos sin
cos sin
Câu 90.x x
I dxx x
42
3 30
cos sin
cos sin
x t dx dt2
Đặt x t dx dt
2
t t x xI dt dx
t t x x
0 4 42
3 3 3 30
2
sin cos sin cos
cos sin cos sin
t t x x
I dt dxt t x x
0 4 42
3 3 3 30
2
sin cos sin cos
cos sin cos sin
x x x x x x x xI dx dx xdx
x x x x
4 4 3 32 2 2
3 3 3 30 0 0
cos sin sin cos sin cos (sin cos ) 1 12 sin2
2 2sin cos sin cos
x x x x x x x xI dx dx xdx
x x x x
4 4 3 32 2 2
3 3 3 30 0 0
cos sin sin cos sin cos (sin cos ) 1 12 sin2
2 2sin cos sin cos
I1
4 I
1
4 .
I x dxx
22
20
1tan (cos )
cos (sin )
Câu 91. I x dx
x
22
20
1tan (cos )
cos (sin )
x t dx dt2
Đặt x t dx dt
2
I t dtt
22
20
1tan (sin )
cos (cos )
x dx
x
22
20
1tan (sin )
cos (cos )
I t dt
t
22
20
1tan (sin )
cos (cos )
x dx
x
22
20
1tan (sin )
cos (cos )
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Trang 20
I x x dxx x
22 2
2 20
1 12 tan (cos ) tan (sin )
cos (sin ) cos (cos )
Do đó: I x x dx
x x
22 2
2 20
1 12 tan (cos ) tan (sin )
cos (sin ) cos (cos )
dt
2
0
2
= dt2
0
2
I2
I
2
.
x xI dx
x
4
0
cos sin
3 sin2
Câu 92.
x xI dx
x
4
0
cos sin
3 sin2
u x xsin cos du
I
u
2
21 4
Đặt u x xsin cos
duI
u
2
21 4
u t2sin
tdtI dt
t
4 4
2
6 6
2cos
124 4sin
. Đặt u t2sin
tdtI dt
t
4 4
2
6 6
2cos
124 4sin
.
xI dx
x x
3
20
sin
cos 3 sin
Câu 93.
xI dx
x x
3
20
sin
cos 3 sin
t x23 sin Đặt t x23 sin x24 cos= x24 cos x t2 2cos 4 . Ta có: 2 x t2cos 4 x x
dt dx
x2
sin cos
3 sin
và x x
dt dx
x2
sin cos
3 sin
.
xdx
x x
3
20
sin.
cos 3 sin
I =
x dx
x x
3
20
sin.
cos 3 sin
x xdx
x x
3
2 20
sin .cos
cos 3 sin
=
x x dx
x x
3
2 20
sin .cos
cos 3 sin
dt
t
15
2
23
4=
dt
t
15
2
23
4 dt
t t
15
2
3
1 1 1
4 2 2
= dt
t t
15
2
3
1 1 1
4 2 2
t
t
15
2
3
1 2ln
4 2
=
t
t
15
2
3
1 2ln
4 2
1 15 4 3 2ln ln
4 15 4 3 2
= 1 15 4 3 2
ln ln4 15 4 3 2
1ln 15 4 ln 3 2
2 = 1 ln 15 4 ln 3 2
2 .
x x x xI dx
x x
2
33 2
3
( sin )sin
sin sin
Câu 94.
x x x xI dx
x x
2
33 2
3
( sin )sin
sin sin
x dxI dx
xx
2 2
3 32
3 31 sinsin
x dxI dx
xx
2 2
3 32
3 31 sinsin
.
xI dx
x
2
31 2
3sin
+ Tính
xI dx
x
2
31 2
3sin
u xdu dx
dxdv v x
x2cot
sin
. Đặt
u x du dx
dxdv v x
x2cot
sin
I13
I1
3
dx dx dxI =
x xx
2 2 2
3 3 32
23 3 3
4 2 31 sin
1 cos 2cos2 4 2
+ Tính dx dx dx
I = x x
x
2 2 2
3 3 32
23 3 3
4 2 31 sin
1 cos 2cos2 4 2
I 4 2 33
Vậy: I 4 2 3
3
.
http://megabook.vn/
Trang 21
xdx
x x
I2
2 20
sin2
cos 4sin
Câu 95.x
dx
x x
I2
2 20
sin2
cos 4sin
x xdx
x
I2
20
2sin cos
3sin 1
x x dx
x
I2
20
2sin cos
3sin 1
u x23sin 1 . Đặt u x23sin 1
ududu
uI
2 2
1 1
22 233 3
udu
duu
I2 2
1 1
22 233 3
x
I dxx
6
0
tan4
cos2
Câu 96.
x
I dxx
6
0
tan4
cos2
xx
I dx dxx x
26 6
20 0
tantan 14
cos2 (tan 1)
x x
I dx dxx x
26 6
20 0
tantan 14
cos2 (tan 1)
t x dt dx x dx
x
2
2
1tan (tan 1)
cos . Đặt t x dt dx x dx
x
2
2
1tan (tan 1)
cos
dtI
tt
11
33
200
1 1 3
1 2( 1)
dtI
tt
11
33
200
1 1 3
1 2( 1)
.
xI dx
x x
3
6
cot
sin .sin4
Câu 97.x
I dx
x x
3
6
cot
sin .sin4
xI dx
x x
3
2
6
cot2
sin (1 cot )
x
I dxx x
3
2
6
cot2
sin (1 cot )
x t1 cot dx dtx2
1
sin . Đặt x t1 cot dx dt
x2
1
sin
t
I dt t tt
3 1 3 1
3 1
3 1 3
3
1 22 2 ln 2 ln 3
3
tI dt t t
t
3 1 3 1
3 1
3 1 3
3
1 22 2 ln 2 ln 3
3
dxI
x x
3
2 4
4
sin .cos
Câu 98.dx
Ix x
3
2 4
4
sin .cos
dxI
x x
3
2 2
4
4.sin 2 .cos
Ta có: dx
Ix x
3
2 2
4
4.sin 2 .cos
dt
t x dxt2
tan1
. Đặt dt
t x dxt2
tan1
t dt tI t dt t
tt t
32 2 33 3(1 ) 1 1 8 3 42( 2 ) ( 2 )2 2 3 3
1 1 1
t dt tI t dt t
tt t
32 2 33 3(1 ) 1 1 8 3 42( 2 ) ( 2 )2 2 3 3
1 1 1
http://megabook.vn/
Trang 22
xI dx
x x x
4
20
sin
5sin .cos 2cos
Câu 99.x
I dxx x x
4
20
sin
5sin .cos 2cos
xI dx
x x x
4
2 20
tan 1.
5tan 2(1 tan ) cos
Ta có: x
I dxx x x
4
2 20
tan 1.
5tan 2(1 tan ) cos
t xtan. Đặt t xtan ,
tI dt dt
t tt t
1 1
20 0
1 2 1 1 2ln3 ln2
3 2 2 1 2 32 5 2
tI dt dt
t tt t
1 1
20 0
1 2 1 1 2ln3 ln2
3 2 2 1 2 32 5 2
xdx
x x xI
24
4 2
4
sin
cos (tan 2tan 5)
Câu 100.
xdx
x x xI
24
4 2
4
sin
cos (tan 2tan 5)
dtt x dx
t2
tan1
Đặt dt
t x dxt2
tan1
t dt dtI
t t t t
21 1
2 21 1
22 ln 3
32 5 2 5
t dt dt
It t t t
21 1
2 21 1
22 ln 3
32 5 2 5
dtI
t t
1
1 21 2 5
Tính dt
It t
1
1 21 2 5
t
u I du0
1
4
1 1tan
2 2 8
. Đặt
t u I du
0
1
4
1 1tan
2 2 8
I
2 32 ln
3 8
. Vậy I
2 32 ln
3 8
.
xI dx
x
22
6
sin
sin3
Câu 101.x
I dxx
22
6
sin
sin3
.
x xI dx dx
x x x
22 2
3 2
6 6
sin sin
3sin 4sin 4cos 1
x x
I dx dxx x x
22 2
3 2
6 6
sin sin
3sin 4sin 4cos 1
t x dt xdxcos sin Đặt t x dt xdxcos sin dt dt
It t
3
0 2
2203
2
1 1ln(2 3)
14 44 14
dt dt
It t
3
0 2
2203
2
1 1ln(2 3)
14 44 14
x xI dx
x
2
4
sin cos
1 sin2
Câu 102.
x xI dx
x
2
4
sin cos
1 sin2
x x x x x1 sin2 sin cos sin cos Ta có: x x x x x1 sin2 sin cos sin cos x ;4 2
(vì x ;4 2
)
x xI dx
x x2
4
sin cos
sin cos
x xI dx
x x2
4
sin cos
sin cos
t x x dt x x dxsin cos (cos sin )
I dt tt
22
11
1 1ln ln2
2
. Đặt t x x dt x x dxsin cos (cos sin )
I dt tt
22
11
1 1ln ln2
2
http://megabook.vn/
Trang 23
I x x xdx2
6 3 5
1
2 1 cos .sin .cos Câu 103. I x x xdx2
6 3 5
1
2 1 cos .sin .cos
t dtt x t x t dt x xdx dx
x x
56 3 6 3 5 2
2
21 cos 1 cos 6 3cos sin
cos sin
t tI t t dt
11 7 13
6 6
00
122 (1 ) 2
7 13 91
Đặt t dt
t x t x t dt x xdx dxx x
56 3 6 3 5 2
2
21 cos 1 cos 6 3cos sin
cos sin
t tI t t dt
11 7 13
6 6
00
122 (1 ) 2
7 13 91
xdxI
x x
4
20
tan
cos 1 cos
Câu 104.
xdxI
x x
4
20
tan
cos 1 cos
xdxI
x x
4
2 20
tan
cos tan 2
Ta có:
xdxI
x x
4
2 20
tan
cos tan 2
2 2 2
2
tan2 tan 2 tan
cos
xt x t x tdt dx
x. Đặt 2 2 2
2
tan 2 tan 2 tan
cos
xt x t x tdt dx
x
3 3
2 2
3 2 tdt
I dtt
3 3
2 2
3 2 tdt
I dtt
xI dx
x x
2
30
cos2
(cos sin 3)
Câu 105.x
I dxx x
2
30
cos2
(cos sin 3)
t x xcos sin 3 Đặt t x xcos sin 3 t
I dtt
4
32
3 1
32
tI dt
t
4
32
3 1
32
.
xI dx
x x
4
2 40
sin4
cos . tan 1
Câu 106.
xI dx
x x
4
2 40
sin4
cos . tan 1
xI dx
x x
4
4 40
sin4
sin cos
Ta có:
xI dx
x x
4
4 40
sin4
sin cos
t x x4 4sin cos . Đặt t x x4 4sin cos I dt
2
2
1
2 2 2 .
xI dx
x
4
20
sin4
1 cos
Câu 107.x
I dxx
4
20
sin4
1 cos
x xI dx
x
24
20
2sin2 (2cos 1)
1 cos
Ta có:
x xI dx
x
24
20
2sin2 (2cos 1)
1 cos
t x2cos . Đặt t x2cos
tI dt
t
1
2
1
2(2 1) 12 6ln
1 3
tI dt
t
1
2
1
2(2 1) 12 6ln
1 3
.
xI dx
x
6
0
tan( )4
cos2
Câu 108.
xI dx
x
6
0
tan( )4
cos2
26
2
0
tan 1
(tan 1)
x
I dxx
Ta có:26
2
0
tan 1
(tan 1)
x
I dxx
t xtan. Đặt t xtan
1
3
2
0
1 3
( 1) 2
dt
It
1
3
2
0
1 3
( 1) 2
dt
It
.
http://megabook.vn/
Trang 24
36
0
tan
cos 2
x
I dxx
Câu 109.
36
0
tan
cos2
x
I dxx
3 36 6tan tan
2 2 2 2cos sin cos (1 tan )0 0
x xI dx dx
x x x x Ta có:
3 36 6tan tan
2 2 2 2cos sin cos (1 tan )0 0
x xI dx dx
x x x x.
t xtanĐặt t xtan
3
33 1 1 2ln
2 6 2 310
tI dt
t
3
33 1 1 2ln
2 6 2 310
tI dt
t.
xI dx
x
2
0
cos
7 cos2
Câu 110.x
I dxx
2
0
cos
7 cos2
xdx
I
x
2
2 20
1 cos
2 6 22 sin
xdxI
x
2
2 20
1 cos
2 6 22 sin
dx
x x
3
4 3 5
4
sin .cos
Câu 111.
dx
x x
3
4 3 5
4
sin .cos
dx
xx
x
3
384
4 3
1
sin.cos
cos
dx
xx
3
24 3
4
1 1.costan
Ta có: dx
x x
x
3
384
4 3
1
sin.cos
cos
dx
xx
3
24 3
4
1 1.costan
.
t xtanĐặt t xtan I t dt
3384
1
4 3 1
I t dt
3384
1
4 3 1
3
2
0
cos cos sin( )
1 cos
x x xI x dx
x
Câu 112.
3
2
0
cos cos sin( )
1 cos
x x xI x dx
x
x x x x xI x dx x x dx dx J K
x x
2
2 20 0 0
cos (1 cos ) sin .sin.cos .
1 cos 1 cos
Ta có:
x x x x xI x dx x x dx dx J K
x x
2
2 20 0 0
cos (1 cos ) sin .sin.cos .
1 cos 1 cos
J x x dx
0
.cos .
+ Tính J x x dx
0
.cos . u x du dx
dv xdx v xcos sin
J 2 . Đặt
u x du dx
dv xdx v xcos sin
J 2
x xK dx
x20
.sin
1 cos
+ Tính x x
K dxx2
0
.sin
1 cos
x t dx dt
t t t t x xK dt dt dx
t t x2 2 20 0 0
( ).sin( ) ( ).sin ( ).sin
1 cos ( ) 1 cos 1 cos
x x x x dx x dxK dx K
x x x2 2 20 0 0
( ).sin sin . sin .2
21 cos 1 cos 1 cos
. Đặt x t dx dt
t t t t x xK dt dt dx
t t x2 2 20 0 0
( ).sin( ) ( ).sin ( ).sin
1 cos ( ) 1 cos 1 cos
x x x x dx x dxK dx K
2 x x x2 20 0 0
( ).sin sin . sin .2
21 cos 1 cos 1 cos
http://megabook.vn/
Trang 25
t xcosdt
Kt
1
21
2 1
Đặt t xcosdt
Kt
1
21
2 1
t u dt u du2tan (1 tan )
u duK du u
u
2 24 44
2
44 4
(1 tan ).
2 2 2 41 tan
, đặt t u dt u du2tan (1 tan )
u duK du u
u
2 24 44
2
44 4
(1 tan ).
2 2 2 41 tan
I2
24
Vậy I
2
24
2
2
6
cosI
sin 3 cos
x
dxx x
Câu 113.
2
2
6
cosI
sin 3 cos
x
dxx x
2
2 2
6
sin cos
sin 3 cos
x x
I dxx x
Ta có: 2
2 2
6
sin cos
sin 3 cos
x x
I dxx x
t x23 cos . Đặt t x23 cos
dtI
t
15
2
23
1ln( 15 4) ln( 3 2)
24
dt
It
15
2
23
1ln( 15 4) ln( 3 2)
24
Dạng 3: Đổi biến số dạng 2
I x x dx2 12sin sin .
2
6
Câu 114. I x x dx2 12sin sin .
2
6
x t t3
cos sin , 02 2
Đặt x t t
3cos sin , 0
2 2
tdt
42
0
3cos
2
I = tdt4
2
0
3cos
2
3 1
2 4 2
=
3 1
2 4 2
.
2
2 2
0
3sin 4cos
3sin 4cos
x x
I dxx x
Câu 115.
2
2 2
0
3sin 4cos
3sin 4cos
x x
I dxx x
2 2 2
2 2 2
0 0 0
3sin 4cos 3sin 4cos
3 cos 3 cos 3 cos
x x x x
I dx dx dxx x x
2 2
2 2
0 0
3sin 4cos
3 cos 4 sin
x xdx dx
x x
2 2 2
2 2 2
0 0 0
3sin 4cos 3sin 4cos
3 cos 3 cos 3 cos
x x x x
I dx dx dxx x x
2 2
2 2
0 0
3sin 4cos
3 cos 4 sin
x xdx dx
x x
2
1 2
0
3sin
3 cos
xI dx
x+ Tính
2
1 2
0
3sin
3 cos
xI dx
xcos sin t x dt xdx. Đặt cos sin t x dt xdx
1
1 2
0
3
3
dt
It
1
1 2
0
3
3
dt
It
http://megabook.vn/
Trang 26
23 tan 3(1 tan ) t u dt u duĐặt 23 tan 3(1 tan ) t u dt u du26
1 2
0
3 3(1 tan ) 3
3(1 tan ) 6
u du
Iu
26
1 2
0
3 3(1 tan ) 3
3(1 tan ) 6
u du
Iu
2
2 2
0
4cos
4 sin
xI dx
x+ Tính
2
2 2
0
4cos
4 sin
xI dx
x1 1sin cos t x dt xdx
1
12 12
10
4ln3
4
dt
I dtt
. Đặt 1 1sin cos t x dt xdx
1
12 12
10
4 ln3
4
dt
I dtt
3ln3
6
IVậy:
3 ln3
6
I
xI dx
x x
4
2
6
tan
cos 1 cos
Câu 116.
xI dx
x x
4
2
6
tan
cos 1 cos
x xI dx dx
x xxx
4 4
2 22
26 6
tan tan
1 cos tan 2cos 1cos
Ta có: x x
I dx dx
x xxx
4 4
2 22
26 6
tan tan
1 cos tan 2cos 1cos
u x du dxx2
1tan
cos Đặt u x du dx
x2
1tan
cos
uI dx
u
1
21
3
2
uI dx
u
1
21
3
2
ut u dt du
u
2
22
2
. Đặt u
t u dt du
u
2
22
2
I dt t3
3
77 33
7 3 73 .
3 3
.
I dt t3
3
77 33
7 3 73 .
3 3
x
I dxx x
2
4
sin4
2sin cos 3
Câu 117.
x
I dxx x
2
4
sin4
2sin cos 3
x xI dx
x x
2
2
4
1 sin cos
2 sin cos 2
Ta có:
x xI dx
x x
2
2
4
1 sin cos
2 sin cos 2
t x xsin cos . Đặt t x xsin cos I dt
t
1
20
1 1
2 2
I dt
t
1
20
1 1
2 2
t u2 tanĐặt t u2 tanu
I duu
1arctan
22
20
1 2(1 tan ) 1 1arctan
22 22tan 2
uI du
u
1arctan
22
20
1 2(1 tan ) 1 1arctan
22 22tan 2
http://megabook.vn/
x xI dx
x
3
2
3
sin
cos
Trang 27
I dxx
3
2
3
cos
Dạng 4: Tích phân từng phần
III dddxxxxxx
333
222
333
cccooosss
x dxI xd J
x x x
3 33
33 3
1 4,
cos cos cos 3
Câu 118. xxx xxx
sssiiinnn
dddxxxxxx
cccooo cccooo
dxJ
x
3
3
cos
.
xxx III ddd JJJ
xxx xxx xxx
333 333333
333333 333
111 444,,,
sss cccooo sss 333
xxx
t xsin .
Sử dụng công thức tích phân từng phần ta có:
sss
dddxxxJJJ
333
333
cccooosss
iiinnn ...d x d t t
Jx tt
3 33 2 2
2 33
23 2
1 1 2 3ln ln
cos 2 1 2 31
với dx
J x
3
ttt xxxsssdx dt t
Jttt
3 33 2 2
222333
222222
1 1 2 3ln ln
333111
I4 2 3
ln .3 2 3
Để tính J ta đặt ddd
xxx
ttt
333 222 222
333
333
111 222lllnnn lll
ccc sss 222 111 222
I 4 222 333
lnnn ...
Khi đó xxx dddttt ttt
JJJ
333 333
111 333nnn
ooo
333 222 333
xxI e dx
x
2
0
1 sin.
1 cos
x xx x
x xx 2 2
1 2sin cos1 sin 12 2 tan1 cos 2
2cos 2cos2 2
Vậy III 444
lll
xxxxxxIII eee dddxxx
222
000
111 sssiiinnn...
cccooo
x xxxx x
xxx 2 2
1 sin coss nnn 111222 2 tan
111 222222 222 sss
222 222
xxe dx x
I e dxx
2 2
20 0
tan2
2cos2
Câu 119. xx
I e dxxxx
2 1 sin.
111 sss
xxx 222 222
222111 iii
ccc ssscccooosss cccooo
eee xxx xxxxxx xxx
III eee dddxxxxxx
222 222
tttaaannn222
e2
Ta có:
xxx xxx xxx
xxx
111 sssiiinnn cccooossssss 222 tttaaannnooo
ddd
222000 000
222 ooosss222
e2
x xI dx
x
4
20
cos2
1 sin2
e x
xdx xI e dx
2 2
tan2
ccc
eee
x 2
0
s
1 sin2
= 222
xxx 222
000
sss
111 sssiiinnn222
Câu 120. xxx xxx
III dddxxx 444 cccooo 222
http://megabook.vn/
Trang 28
u x du dxx
dv dx vxx 2
cos2 1
1 sin2(1 sin2 )
Đặt
u x du dxx
dv dx v xx 2
cos2 1
1 sin2(1 sin2 )
I x dx dxx x
x
4 4
20 0
1 1 1 1 1 1 1. . .4
2 1 sin2 2 1 sin2 16 2 20 cos4
x1 1 1 2 2
. tan . 0 1416 2 4 16 2 2 4 162 0
I x dx dxx x
x
4 4
20 0
1 1 1 1 1 1 1. . .4
2 1 sin2 2 1 sin2 16 2 20 cos4
x1 1 1 2 2
. tan . 0 1416 2 4 16 2 2 4 162 0
TP4: TÍCH PHÂN HÀM SỐ MŨ - LOGARIT
Dạng 1: Đổi biến số
x
x
eI dx
e
2
1
Câu 121.
x
x
eI dx
e
2
1
x x xt e e t e dx tdt2 2 Đặt x x xt e e t e dx tdt2 2
tI dt
t
3
21
t t t t C3 222 2ln 1
3 x x x x xe e e e e C
22 2ln 1
3
.
tI dt
t
3
21
t t t t C3 222 2ln 1
3 x x x x xe e e e e C
22 2ln 1
3
x
x
x x eI dx
x e
2( )
Câu 122.
x
x
x x eI dx
x e
2( )
x
x
x x eI dx
x e
2( )
x
x
x x eI dx
x e
2( )
x x
x
xe x edx
xe
.( 1)
1
=
x x
x
xe x e dx
xe
.( 1)
1
xt x e. 1 . Đặt xt x e. 1 x xI xe xe C1 ln 1 x xI xe xe C1 ln 1 .
x
dxI
e2 9
Câu 123.
x
dxI
e2 9
xt e2 9 Đặt xt e2 9 dt t
I Ctt2
1 3ln
6 39
x
x
eC
e
2
2
1 9 3ln
6 9 3
dt t
I Ctt2
1 3ln
6 39
x
x
e C
e
2
2
1 9 3ln
6 9 3
x
x
x xI dx
ex e2
2
2 1
ln(1 ) 2011
ln ( )
Câu 124.
x
x
x xI dx
ex e2
2
2 1
ln(1 ) 2011
ln ( )
x xI dx
x x
2
2 2
ln( 1) 2011
( 1) ln( 1) 1
Ta có: x x
I dxx x
2
2 2
ln( 1) 2011
( 1) ln( 1) 1
t x2ln( 1) 1 . Đặt t x2ln( 1) 1
tI dt
t
1 2010
2
t t C
11005ln
2
tI dt
t
1 2010
2
t t C
11005ln
2 x x C2 21 1
ln( 1) 1005ln(ln( 1) 1)2 2
= x x C2 21 1ln( 1) 1005ln(ln( 1) 1)
2 2
http://megabook.vn/
Trang 29
e x
x
xeJ dx
x e x1
1
( ln )
Câu 125.
e x
x
xeJ dx
x e x1
1
( ln )
e x eex
x
d e x eJ e x
ee x 11
( ln ) 1ln ln ln
ln
e x ee
x
x
d e x eJ e x
ee x 11
( ln ) 1ln ln ln
ln
x x
x x x
e eI dx
e e e
ln2 3 2
3 20
2 1
1
Câu 126.
x x
x x x
e eI dx
e e e
ln2 3 2
3 20
2 1
1
x x x x x x
x x x
e e e e e eI dx
e e e
ln2 3 2 3 2
3 20
3 2 ( 1)
1
x x x x x x
x x x
e e e e e eI dx
e e e
ln2 3 2 3 2
3 20
3 2 ( 1)
1
x x x
x x x
e e edx
e e e
ln2 3 2
3 20
3 21
1
=
x x x
x x x
e e e dx
e e e
ln2 3 2
3 20
3 21
1
x x xe e e x3 2 ln2 ln2ln( – 1)
0 0 =
x x xe e e x3 2 ln2 ln2ln( – 1)
0 0
14ln
4= ln11 – ln4 =
14ln
4
x
dxI
e
3ln2
230 2
Câu 127.
x
dxI
e
3ln2
230 2
x
xx
e dxI
e e
3ln2 3
20 33 2
x
xx
e dxI
e e
3ln2 3
20 33 2
x x
t e dt e dx3 31
3 . Đặt
x x
t e dt e dx3 31
3 I
3 3 1ln
4 2 6
I
3 3 1ln
4 2 6
xI e dxln2
3
0
1 Câu 128.xI e dx
ln23
0
1
xe t3
1 t dt
dxt
2
3
3
1
t dtdx
t
2
3
3
1
dt
t
1
30
13 1
1
I = dt
t
1
30
13 1
1
dt
t
1
30
3 31
= dt
t
1
30
3 31
.
dtI
t
1
1 30
31
Tính dt
It
1
1 30
31
t
dtt t t
1
20
1 2
1 1
=
t dt
t t t
1
20
1 2
1 1
ln2
3
= ln2
3
I 3 ln23
Vậy: I 3 ln2
3
x x
x x x x
e e dxI
e e e e
ln15 2
3ln2
24
1 5 3 1 15
Câu 129.
x x
x x x x
e e dxI
e e e e
ln15 2
3ln2
24
1 5 3 1 15
x xt e t e21 1 xe dx tdt2 Đặt x xt e t e21 1 xe dx tdt2
t t dtI dt t t t
t tt
4 42 4
2 33 3
(2 10 ) 3 72 2 3ln 2 7ln 2
2 24
2 3ln2 7ln6 7ln5
.
t t dtI dt t t t
t tt
4 42 4
2 33 3
(2 10 ) 3 72 2 3ln 2 7ln 2
2 24
2 3ln2 7ln6 7ln5 ln3 2
ln 2 1 2
x
x x
e dxI
e e
Câu 130.
ln3 2
ln2 1 2
x
x x
e dxI
e e
xe 2 xe dx tdt2 2 Đặt t = xe 2 xe dx tdt2 2
t tdt
t t
1 2
20
( 2)
1
I = 2
t tdt
t t
1 2
20
( 2)
1
tt dt
t t
1
20
2 11
1
= 2
tt dt
t t
1
20
2 11
1
t dt
1
0
2 ( 1)= 1
t dt
0
2 ( 1)d t t
t t
1 2
20
( 1)2
1
+
d t t
t t
1 2
20
( 1)2
1
t t1
20
( 2 )= t t1
20
( 2 ) t t1
20
2ln( 1) + t t1
20
2ln( 1) 2ln3 1= 2ln3 1 .
http://megabook.vn/
Trang 30
x x
x x
e eI dx
e e
ln3 3 2
0
2
4 3 1
Câu 131.
x x
x x
e eI dx
e e
ln3 3 2
0
2
4 3 1
x x x x x xt e e t e e tdt e e dx3 2 2 3 2 3 24 3 4 3 2 (12 6 ) x x tdte e dx3 2(2 )
3
tdtI dt
t t
9 9
1 1
1 1 1(1 )
3 1 3 1
t t 9
1
1 8 ln5( ln 1) .
3 3
Đặt x x x x x xt e e t e e tdt e e dx3 2 2 3 2 3 24 3 4 3 2 (12 6 ) x x tdte e dx3 2 (2 )
3
tdtI dt
t t
9 9
1 1
1 1 1(1 )
3 1 3 1
t t 9
1
1 8 ln5( ln 1) .
3 3
3
16ln
3
8ln
43 dxeI xCâu 132.
3
16ln
3
8ln
43 dxeI x
x x tt e e
2 43 4
3
tdtdx
t22
4
t dtI dt dt
t t
2 3 2 3 2 32
2 22 2 2
22 8
4 4
I14 3 1 8
Đặt: x x tt e e
2 43 4
3
tdtdx
t22
4
t dtI dt dt
t t
2 3 2 3 2 32
2 22 2 2
22 8
4 4
4 I13 1 8
dtI
t
2 3
1 22 4
, với dt
It
2 3
1 2
2 4
dtI
t
2 3
1 22 4
Tính dt
It
2 3
1 2
2 4 t u u2tan , ;
2 2
dt u du22(1 tan )
I du3
1
4
1 1
2 2 3 4 24
. Đặt: t u u2tan , ;2 2
dt u du22(1 tan )
I du3
1
4
1 1
2 2 3 4 24
I 4( 3 1)
3
. Vậy: I 4( 3 1)
3
x
x
eI dx
e
ln3
30 ( 1)
Câu 133.
x
x
eI dx
e
ln3
30 ( 1)
x x x
x
tdtt e t e tdt e dx dx
e
2 21 1 2
tdtI
t
2
32
2 2 1 Đặt x x x
x
tdtt e t e tdt e dx dx
e
2 21 1 2
tdtI
t
2
32
2 2 1
x
x
eI dx
e
ln5 2
ln2 1
Câu 134.
x
x
eI dx
e
ln5 2
ln2 1
x x
x
tdt tt e t e dx I t d t
e
22 3
2 2
11
2 201 1 2 ( 1) 2
3 3
Đặt x x
x
tdt tt e t e dx I t d t
e
22 3
2 2
11
2 201 1 2 ( 1) 2
3 3
xI e dxln2
0
1 Câu 135.xI e dx
ln2
0
1
x x x
x
td tdt e t e tdt e dx dx
e t
2
2
2 21 1 2
1
tI dt dt
t t
1 12
2 20 0
2 1 42 1
21 1
http://megabook.vn/
Trang 31
x x
x xI dx
2
1
2 2
4 4 2
Câu 136.
x x
x xI dx
2
1
2 2
4 4 2
x xt 2 2 Đặt x xt 2 2 x x x x 24 4 2 (2 2 ) 4 x x x x 24 4 2 (2 2 ) 4 1 81
ln4ln 2 25
I1 81
lnI 4ln 2 25
1
0
6
9 3.6 2.4
x
x x x
dxICâu 137.
1
0
6
9 3.6 2.4
x
x x x
dxI
x
x x
dx
I1
20
3
2
3 33 2
2 2
Ta có:
x
x x
dx
I1
20
3
2
3 33 2
2 2
x
t3
2
. Đăt
x
t 3
2
dtI
t t
3
2
21
1
ln3 ln2 3 2
ln15 ln14
ln3 ln2
.
dtI
t t
3
2
21
1
ln3 ln2 3 2
ln15 ln14
ln3 ln2
ex
I x x dxx x
2
1
ln3 ln
1 ln
Câu 138.
e x
I x x dxx x
2
1
ln3 ln
1 ln
e ex
I dx x xdxx x
2
1 1
ln3 ln
1 ln
e e
xI dx x xdx
x x
2
1 1
ln3 ln
1 ln
2(2 2)
3
=
2(2 2)
3
e32 1
3
+
e32 1
3
e35 2 2 2
3
=
e35 2 2 2
3
ex x
I dxx
3 2
1
ln 2 ln Câu 139.
e x x
I dxx
3 2
1
ln 2 ln
t x22 ln Đặt t x22 ln x
dt dxx
2ln
xdt dx
x
2ln I tdt
33
2
1
2 33 4 43
3 28
I tdt3
3
2
1
2 33 4 43
3 28
e
e
dxI
x x ex
2
ln .ln Câu 140.
e
e
dxI
x x ex
2
ln .ln
e e
e e
dx d xI
x x x x x
2 2
(ln )
ln (1 ln ) ln (1 ln )
e e
e e
dx d xI
x x x x x
2 2
(ln )
ln (1 ln ) ln (1 ln )
e
e
d xx x
2
1 1(ln )
ln 1 ln
=
e
e
d xx x
2
1 1(ln )
ln 1 ln
= 2ln2 – ln3
x
x x
eI dx
e e
ln6 2
ln4 6 5
Câu 141.
x
x x
eI dx
e e
ln6 2
ln4 6 5
xt e Đặt xt e I 2 9ln3 4ln2 . I 2 9ln3 4ln2
e xI dx
x x
32
21
log
1 3ln
Câu 142.
e xI dx
x x
32
21
log
1 3ln
e e e
x
x x xdxI dx dx
xx x x x x
3
3 22
32 2 21 1 1
ln
log ln2 1 ln . ln.
ln 21 3ln 1 3ln 1 3ln
e e e
x
x x xdxI dx dx
xx x x x x
3
3 22
32 2 21 1 1
ln
log ln2 1 ln . ln.
ln 21 3ln 1 3ln 1 3ln
dxx t x t x tdt
x
2 2 21 11 3ln ln ( 1) ln .
3 3 Đặt
dxx t x t x tdt
x
2 2 21 11 3ln ln ( 1) ln .
3 3 .
I t t
2
3
3 31
1 1 4
39ln 2 27ln 2
Suy ra I t t
2
3
3 31
1 1 4
39ln 2 27ln 2
.
http://megabook.vn/
Trang 32
ex x x
I dxx x
1
( 2) ln
(1 ln )
Câu 143.
e x x x
I dxx x
1
( 2) ln
(1 ln )
e ex
dx dxx x
1 1
ln2
(1 ln )
e e
xdx dx
x x1 1
ln2
(1 ln )
ex
e dxx x
1
ln1 2
(1 ln )
= e
xe dx
x x1
ln1 2
(1 ln )
ex
dxx x
1
ln
(1 ln )Tính J = e
x dx
x x1
ln
(1 ln ) t x1 ln . Đặt t x1 ln
tJ dt
t
2
1
11 ln2
tJ dt
t
2
1
11 ln2 .
I e 3 2ln2 Vậy: I e 3 2ln2 .
e
e
x x x xI dx
x x
3
2
2 22 ln ln 3
(1 ln )
Câu 144.
e
e
x x x xI dx
x x
3
2
2 22 ln ln 3
(1 ln )
e e
e e
I dx xdxx x
3 3
2 2
13 2 ln
(1 ln )
e e3 23ln2 4 2 e e
e e
I dx xdxx x
3 3
2 2
13 2 ln
(1 ln )
e e3 23ln2 4 2 .
ex x
I dxx
22 2
21
ln ln 1 Câu 145.
e x x
I dxx
22 2
21
ln ln 1
dxt x dt
xln Đặt :
dxt x dt
xln
t t t t
t t t t tI dt dt dt dt I I
e e e e
22 2 1 2
1 20 0 0 1
2 1 1 1 1
t t t t
t t t t tI dt dt dt dt I I
e e e e
22 2 1 2
1 20 0 0 1
2 1 1 1 1
t
t t t t
tdt dt dt dtI te
ee e e e
11 1 1 1
1 0 0 0 00
1
+ t
t t t t
tdt dt dt dtI te
ee e e e
11 1 1 1
1 0 0 0 00
1
t t
t t t t
tdt dt dt dtI te te
ee e e e e
2 22 2 2 2
2 1 1 1 1 21 1
1 2 + t t
t t t t
tdt dt dt dtI te te
ee e e e e
2 22 2 2 2
2 1 1 1 1 21 1
1 2
eI
e2
2( 1)Vậy :
eI
e2
2( 1)
5
2
ln( 1 1)
1 1
xI dx
x xCâu 146.
5
2
ln( 1 1)
1 1
xI dx
x x
t xln 1 1 Đặt t xln 1 1 dx
dtx x
21 1
dx
dtx x
21 1
I dtln3
2 2
ln2
2 ln 3 ln 2 I dtln3
2 2
ln2
2 ln 3 ln 2 .
33
1
ln
1 ln
e
xI dx
x xCâu 147.
3 3
1
ln
1 ln
e
xI dx
x x
dxt x x t tdt
x
21 ln 1 ln 2 Đặt dx
t x x t tdtx
21 ln 1 ln 2 x t3 2 3ln ( 1) và 3 x t2 3ln ( 1)
t t t tI dt = dt t t t dt
t t t
2 2 22 3 6 4 25 3
1 1 1
( 1) 3 3 1 1( 3 3 )
15ln2
4
t t t tI dt = dt t t t dt
t t t
2 2 22 3 6 4 25 3
1 1 1
( 1) 3 3 1 1( 3 3 )
15ln2
4
ex
I dxx x1
3 2ln
1 2ln
Câu 148.
e x
I dxx x1
3 2ln
1 2ln
t x1 2ln Đặt t x1 2ln
e
I t dt2
1
(2 ) e
I t dt2
1
(2 ) 3
524 =
3
524
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Trang 33
ex x
I dxx
3 2
1
ln 2 ln Câu 149.
e x x
I dxx
3 2
1
ln 2 ln t x22 ln Đặt t x22 ln I
33 4 433 2
8
I
33 4 433 2
8
1
1
( ln )
e x
x
xeI dx
x e xCâu 150.
1
1
( ln )
e x
x
xeI dx
x e x
xt e xln Đặt xt e xln 1
ln
ee
Ie
1
ln
ee
Ie
.
Dạng 2: Tích phân từng phần
inxI e xdx2
s
0
.sin2
Câu 151.inxI e xdx
2s
0
.sin2
inxI e x xdx2
s
0
2 .sin cos
inxI e x xdx2
s
0
2 .sin cos
x x
u x du xdx
dv e xdx v esin sin
sin cos
cos
. Đặt x x
u x du xdx
dv e xdx v esin sin
sin cos
cos
x x xI xe e xdx e e2
sin sin sin2 20 0
0
2sin .cos 2 2 2
I x x x dx1
2
0
ln( 1) Câu 152. I x x x dx1
2
0
ln( 1)
http://megabook.vn/
xdu dx
u x x x xdv xdx x
v
2 2
2
2 1
ln( 1) 1
2
x x xI x x dx
x x
112 3 2
2
20 0
1 2ln( 1)
2 2 1
x dxx dx dx
x x x x
1 1 1
2 20 0 0
1 1 1 2 1 3ln3 (2 1)
2 2 4 41 1
3 3ln3
4 12
Trang 34
xxxddduuu dddxxx
uuu xxx xxx xxx xxxdv xdx x
v
222
2
222 111
lllnnn((( 111))) 111
2
x x xI x x dx
x x
112 3 2
2
20 0
1 2ln( 1)
2 2 1
x dxx dx dx
x x x x
1 1 1
2 20 0 0
1 1 1 2 1 3ln3 (2 1)
2 2 4 41 1
3 3ln3
4 12
xI dx
x
8
3
ln
1
Đặt
dv xdx x
v
222
2
2
x x x I x x dx
x x
112 3 2
2
20 0
1 2ln( 1)
2 2 1
x dx x dx dx
x x x x
1 1 1
2 20 0 0
1 1 1 2 1 3ln3 (2 1)
2 2 4 41 1
3 3ln3
4 12
xI dx
xxx
8ln
111
u x dxdu
dx xdvv xx
ln
2 11
xI x x dx J
x
88
33
12 1.ln 2 6ln8 4ln3 2
Câu 153. x
I dx 8
333
ln
uuu xxx dddddd
ddddv
v xx 2 11
xxxxxx xxx ddd
x
888888
3
111222 111...lllnnn 222 666lllnnn888 444lllnnn333 222
xJ dx
x
8
3
1
Đặt
xxx uuu
xxx xxx dv v x x
lllnnn
2 11
III xxx JJJ x 333
3
dddx
888
3
t tt x J tdt dt dt
t tt t
3 3 32
2 22 2 2
1 11 .2 2 2
1 11 1
tt
t
83
12 ln 2 ln3 ln2
1
+ Tính xxx
JJJ xxx x
3
111 ddd ddd
t tt t
333 333 333222
2 22 2 2
... 2221 1
tt
t
83
12 ln 2 ln3 ln2
1
I 20ln2 6ln3 4
. Đặt ttt ttt
ttt xxx JJJ ttt ttt ttt dddttt t t t t 2 2
2 2 2
111 111111 222 222
111 1111 1
t t
t
83
12 ln 2 ln3 ln2
1
I 20ln2 6ln3 4 Từ đó I 20ln2 6ln3 4
exx x x
I e dxx
2
1
ln 1
.
eee xxxxxx xxx xxx
eee dddxxxx
222
1
lllnnn 111
e e e xx x e
I xe dx xe dx dxx
1 1 1
ln
Câu 154.
III x
1
eee eee eee xxxxxx xxx eee
III eee xxx xxxeee dddxxx xxxx
1 1 1
lll e ee
x x x eI xe dx xe e dx e e111 1
( 1)
xxx ddd ddd x
1 1 1
nnn eee eeeeee
xxx xxx xxx eeeddd xxx 1111111 1
e e ex xex x ee e
I e xdx e x dx e dxx x2
11 1 1
ln ln
. + Tính
III xxxeee xxx eee eee dddxxx eee eee
1 1
((( 111)))
eee xxx xxxxxx eee eee
III eee xxx eee xxx dddxxx eee xxxx x222
1111 1 1
+Tính eee eee eee
xxx eee xxxddd ddd
x x 1 1 1
lllnnn lllnnn
e xeI I I dx
x1 21
.
eee xxx
III III xxxx111 222
1
ee 1Vậy:
eee III ddd
x 1
eeeeee 111=
ex
I x dxx x
2
1
lnln
1 ln
.
eee
III dddxxxx x
222
1
lll1 ln
ex
I dxx x
11
ln
1 ln
Câu 155.
xxx
xxx x x 1
lllnnnnnn
1 ln
eee
x x111
1 1 ln t x1 ln Tính
xxx III dddxxx
x x1
lllnnn
1 ln
xxx111 I1
4 2 2
3 3 . Đặt ttt lllnnn III111
444 222 222
3 3
3 3
e
I xdx22
1
ln
.
eee
III xxxdddxxx222222
1
nnn I e2 2 + Tính
1
lll 222. Lấy tích phân từng phần 2 lần được III eee 222
I e2 2 2
3 3
.
III eee 222 222 222
3 3 Vậy
3 3 .
http://megabook.vn/
Trang 35
2
3
2
1
ln( 1)xI dx
x
Câu 156.
2
3
2
1
ln( 1)xI dx
x
xduu x
xdxdv
vxx
22
32
2ln( 1)
11
2
Đặt
xduu x
xdxdv
vx x
22
32
2ln( 1)
11
2
x dx
x xx
22
2 21
2ln( 1)
12 ( 1)
xdx
x x
2
21
ln2 ln5 1
2 8 1
dx dx
x x
2 2 2
21 1
ln2 ln5 1 ( 1)
2 8 2 1
x x2 2ln2 ln5 1ln | | ln | 1|
2 8 2 1
. Do đó I = x dx
x x x
22
2 21
2ln( 1)
12 ( 1)
x dx
x x
2
21
ln2 ln5 1
2 8 1
dx d x
x x
2 2 2
21 1
ln2 ln5 1 ( 1)
2 8 2 1
x x2 2ln2 ln5 1ln | | ln | 1|
2 8 2 1
52ln2 ln5
8=
52ln2 ln5
8
xI = dx
x
2
21
ln( 1)Câu 157.
xI = dx
x
2
21
ln( 1)
dxu x du dxxdx I xdv x x x
vxx
2
2 1
ln( 1)1 321 ln( 1) 3ln2 ln3
1 1 ( 1) 2
Đặt
dxu x du dxxdx I xdv x x x
vx x
2
2 1
ln( 1)1 321 ln( 1) 3ln2 ln3
1 1 ( 1) 2
xI x dx
x
1
2
0
1ln
1
Câu 158.
xI x dx
x
1
2
0
1ln
1
du dxxu x
xxdv xdx v
2
2
21
ln (1 )1
2
Đặt
du dxxu x
x xdv xdx v
2
2
21
ln (1 )1
2
xI x x dx
x x
1
22 2
20
11 1 2
ln 22 1 10
xdx dx
x xx
1 1
22 2
20 0
ln3 ln3 1 ln3 1 1 21 ln
8 8 ( 1)( 1) 8 2 2 31
x
I x x dxx x
1
22 2
20
11 1 2
ln 22 1 10
x dx dx
x xx
1 1
22 2
20 0
ln3 ln3 1 ln3 1 1 21 ln
8 8 ( 1)( 1) 8 2 2 31
I x x dxx
22
1
1.ln
Câu 159. I x x dx
x
22
1
1.ln
u xx
dv x dx2
1ln
Đặt u x
x
dv x dx2
1ln
I10 1
3ln3 ln23 6
I10 1
3ln3 ln23 6
I x x dx1
2 2.ln(1 )
0
Câu 160. I x x dx1
2 2.ln(1 )
0
u x
dv x dx
2
2
ln(1 )
Đặt u x
dv x dx
2
2
ln(1 )
I1 4
.ln23 9 6
I
1 4.ln2
3 9 6
xI dx
x
3
21
ln
( 1)
Câu 161.
xI dx
x
3
21
ln
( 1)
u x
dxdv
x 2
ln
( 1)
Đặt
u x
dxdv
x 2
ln
( 1)
I1 3
ln3 ln4 2
I 1 3
ln3 ln4 2
http://megabook.vn/
Trang 36
2 2
1
ln ( ln ).
1
e x x
x
x e e xI dx
eCâu 162.
2 2
1
ln ( ln ).
1
e x x
x
x e e xI dx
e
e e x
x
eI x dx dx H K
e
22
1 1
ln .1
Ta có: e e x
x
eI x dx dx H K
e
22
1 1
ln .1
e
H x dx2
1
ln . + e
H x dx2
1
ln . u x
dv dx
2ln
. Đặt: u x
dv dx
2ln
e
H e x dx e
1
2ln . 2 e
H e x dx e
1
2ln . 2
e x
x
eK dx
e
2
1 1
+ e x
x
eK dx
e
2
1 1
xt e 1 . Đặt xt e 1
eee
ee
t eI dt e e
t e
1
21
1 1ln
1
ee e
ee
t eI dt e e
t e
1
21
1 1ln
1
e
e
eI e
e
1–2 ln
1
Vậy: e
e
eI e
e
1–2 ln
1
2 1
1
2
1( 1 )
x
xI x e dxx
Câu 163.
2 1
1
2
1( 1 )
x
xI x e dxx
2 31 1
1 1
2 2
1
x x
x xI e dx x e dx H Kx
Ta có:
2 31 1
1 1
2 2
1
x x
x xI e dx x e dx H Kx
2 21 1 5
2
1 12 2
1 3
2
x x
x xH xe x e dx e Kx
5
23
.2
I e
+ Tính H theo phương pháp từng phần I1 =
2 21 1 5
2
1 12 2
1 3
2
x x
x xH xe x e dx e Kx
5
23
.2
I e
4
2
0
ln( 9 ) I x x dxCâu 164.
4
2
0
ln( 9 ) I x x dx
u x x
dv dx
2ln 9
Đặt u x x
dv dx
2ln 9
xI x x x dx
x
4 42
20 0
ln 9 2
9
x
I x x x dx
x
4 42
20 0
ln 9 2
9
http://megabook.vn/
Trang 37
TP5: TÍCH PHÂN TỔ HỢP NHIỀU HÀM SỐ
x xI x e dx
x
31 4
2
0 1
Câu 165.
x xI x e dx
x
31 4
2
0 1
x xI x e dx dx
x
31 1 4
2
0 01
x x
I x e dx dxx
31 1 4
2
0 01
.
xI x e dx3
12
10
+ Tính xI x e dx3
12
10
t x3. Đặt t x3 t tI e dt e e1 1
10
0
1 1 1 1
3 3 3 3 t tI e dt e e
1 1
10
0
1 1 1 1
3 3 3 3 .
xI dx
x
1 4
201
+ Tính x
I dxx
1 4
201
t x4. Đặt t x4t
I dtt
1 4
2 20
24 4
3 41
tI dt
t
1 4
2 20
24 4
3 41
I e1
33
Vậy: I e1
33
x xI x e dx
x
2 2
31
4
Câu 166.x x
I x e dxx
2 2
31
4
xI xe dx2
1
xI xe dx2
1
x
dxx
2 2
21
4+
x dx
x
2 2
21
4 .
xI xe dx e2
21
1
+ Tính xI xe dx e2
21
1
x
I dxx
2 2
2 21
4 + Tính
xI dx
x
2 2
2 21
4 x t2sin. Đặt x t2sin t 0;
2
, t 0;2
.
tI dt t t
t
222
2 2
66
cos( cot )
sin
t
I dt t tt
222
2 2
66
cos( cot )
sin
33
= 3
3
I e2 33
Vậy: I e2 3
3
.
xxI e x x dx
x
12 2 2
20
. 4 .
4
Câu 167. xx
I e x x dx
x
12 2 2
20
. 4 .
4
x xI xe dx dx I I
x
1 1 32
1 22
0 0 4
x x
I xe dx dx I I
x
1 1 32
1 22
0 0 4
x eI xe dx
1 22
10
1
4
+ Tính x e
I xe dx1 2
21
0
1
4
xI dx
x
1 3
22
0 4
+ Tính
xI dx
x
1 3
22
0 4
t x24 . Đặt t x24 I2
163 3
3 I2
16 3 3
3
eI
2 613 3
4 12
eI
2 613 3
4 12
http://megabook.vn/
Trang 38
xxI e dx
x
1 2
20
1
( 1)
Câu 168.
xxI e dx
x
1 2
20
1
( 1)
t x dx dt1 t tt tI e dt e dt
tt t
2 221 1
2 21 1
2 2 2 21
Đặt t x dx dt1 t tt tI e dt e dt
tt t
2 221 1
2 21 1
2 2 2 21
e
e ee
221 1
2
= e
e ee
221 1
2
xx e dxI
x
23 3 1
20
.
1
Câu 169.
xx e dxI
x
23 3 1
20
.
1
t x dx tdt21 Đặt t x dx tdt21 tI t e dt2
2
1
( 1) t tt e dt e J e e
22 2
1
2( )
1 tI t e dt
22
1
( 1) t t
2
t e dt e J e e2 2
1
2( )
1
t t t t t t tJ t e dt t e te dt e e te e dt e e te e2 2 2
2 2 2 2
1 1 1
2 2 22 4 2 4 2( )
1 1 1
+ t t t t t t tJ t e dt t e te dt e e te e dt e e te e2 2 2
2 2 2 2
1 1 1
2 2 22 4 2 4 2( )
1 1 1
I e2Vậy: I e2
x x xI dx
x
2 3
2
ln( 1)
1
Câu 170.
x x xI dx
x
2 3
2
ln( 1)
1
x x x x x x x xf x x
x x x x
2 2 2
2 2 2 2
ln( 1) ( 1) ln( 1)( )
1 1 1 1
Ta có:
x x x x x x x xf x x
x x x x
2 2 2
2 2 2 2
ln( 1) ( 1) ln( 1)( )
1 1 1 1
F x f x dx x d x xdx d x2 2 21 1( ) ( ) ln( 1) ( 1) ln( 1)
2 2 F x f x dx x d x xdx d x2 2 21 1
( ) ( ) ln( 1) ( 1) ln( 1)2 2
x x x C2 2 2 21 1 1ln ( 1) ln( 1)
4 2 2 = 2 x x x C2 2 21 1 1
ln ( 1) ln( 1)4 2 2
.
x x xI dx
x
4 2 3
20
ln 9 3
9
Câu 171.
x x xI dx
x
4 2 3
20
ln 9 3
9
x x x x x xI dx dx dx I I
x x x
4 4 42 3 2 3
1 22 2 2
0 0 0
ln 9 3 ln 93 3
9 9 9
x x x x x xI dx dx dx I I
x x x
4 4 42 3 2 3
1 22 2 2
0 0 0
ln 9 3 ln 93 3
9 9 9
x xI dx
x
4 2
12
0
ln 9
9
+ Tính
x xI dx
x
4 2
12
0
ln 9
9
x x u2ln 9 . Đặt x x u2ln 9 du dx
x2
1
9
du dx
x2
1
9
uI udu
ln9 2 2 2
1ln3
ln 9 ln 3ln9
ln32 2
uI udu
ln9 2 2 2
1ln3
ln 9 ln 3ln9
ln32 2
xI dx
x
4 3
22
0 9
+ Tính
xI dx
x
4 3
22
0 9
x v2 9 . Đặt x v2 9
xdv dx x v
x
2 2
2, 9
9
x
dv dx x v
x
2 2
2, 9
9
uI u du u
5 32
23
445( 9) ( 9 )
33 3
uI u du u
5 32
23
445( 9) ( 9 )
33 3
x x xI dx I I
x
4 2 3 2 2
1 22
0
ln 9 3 ln 9 ln 33 44
29
Vậy
x x xI dx I I
x
4 2 3 2 2
1 22
0
ln 9 3 ln 9 ln 33 44
29
.
http://megabook.vn/
Trang 39
ex x x
I dxx x
3 2
1
( 1) ln 2 1
2 ln
Câu 172.
e x x x
I dxx x
3 2
1
( 1) ln 2 1
2 ln
e ex
I x dx dxx x
2
1 1
1 ln
2 ln
e e
xI x dx dx
x x
2
1 1
1 ln
2 ln
ee
x ex dx
3 32
11
1
3 3
. +
ee
x ex dx
3 32
11
1
3 3
e eex d x x
dx x xx x x x
1
1 1
1 ln (2 ln )ln 2 ln
2 ln 2 ln
e 2ln
2
+
e e ex d x x
dx x xx x x x
1
1 1
1 ln (2 ln )ln 2 ln
2 ln 2 ln
e 2ln
2
e eI
3 1 2ln
3 2
. Vậy:
e eI
3 1 2ln
3 2
.
ex
I dxx x
33
1
ln
1 ln
Câu 173.
e x
I dxx x
33
1
ln
1 ln
dxt x x t tdt
x
21 ln 1 ln 2 Đặt dx
t x x t tdtx
21 ln 1 ln 2 x t3 2 3ln ( 1) và 3 x t2 3ln ( 1)
t t t tI dt = dt t t t dt
t t t
2 2 22 3 6 4 25 3
1 1 1
( 1) 3 3 1 1( 3 3 )
15ln2
4
t t t tI dt = dt t t t dt
t t t
2 2 22 3 6 4 25 3
1 1 1
( 1) 3 3 1 1( 3 3 )
15ln2
4
4
2
0
sin
cos
x x
I dxx
Câu 174.
4
2
0
sin
cos
x x
I dxx
u x du dx
xdv dx v
xx2
sin 1
coscos
Đặt
u x du dx
xdv dx v
xx2
sin 1
coscos
x dx dxI
x x x
4 44
0 0 0
2
cos cos 4 cos
x dx dxI
x x x
4 44
0 0 0
2
cos cos 4 cos
dx xdxI
x x
4 4
1 20 0
cos
cos 1 sin
+ dx xdx
I x x
4 4
1 20 0
cos
cos 1 sin
t xsin. Đặt t xsindt
It
2
2
1 20
1 2 2ln
2 2 21
dtI
t
2
2
1 20
1 2 2ln
2 2 21
2 1 2 2ln
4 2 2 2
Vậy:
2 1 2 2ln
4 2 2 2
4 3
2
1
ln(5 ) . 5
x x xI dx
xCâu 175.
4 3
2
1
ln(5 ) . 5
x x xI dx
x4 4
2
1 1
ln(5 )5 .
xI dx x x dx K H
x Ta có:
4 4
2
1 1
ln(5 ) 5 .
xI dx x x dx K H
x.
xK dx
x
4
21
ln(5 ) +
xK dx
x
4
21
ln(5 )
u x
dxdv
x2
ln(5 )
. Đặt
u x
dxdv
x2
ln(5 )
K3
ln45
K 3
ln45
x x dx4
1
5 .+ H=4
x x dx
1
5 . t x5 . Đặt t x5 H164
15 H
164
15
I3 164
ln45 15
Vậy: I 3 164
ln45 15
I x x x dx
0
22(2 ) ln(4 ) Câu 176. I x x x dx
0
22(2 ) ln(4 )
http://megabook.vn/
Trang 40
I x x dx2
0
(2 ) Ta có: I x x dx2
0
(2 ) x dx2
2
0
ln(4 )+ x dx2
2
0
ln(4 ) I I1 2= I I1 2
I x x dx x dx2 2
21
0 0
(2 ) 1 ( 1)2
+ I x x dx x dx
2 22
10 0
(2 ) 1 ( 1)2
x t1 sin (sử dụng đổi biến: x t1 sin )
xI x dx x x dx
x
2 2 222 2
2 0 20 0
ln(4 ) ln(4 ) 24
+ x
I x dx x x dxx
2 2 222 2
2 0 20 0
ln(4 ) ln(4 ) 24
(sử dụng tích phân từng phần)
6ln2 4 x t2tan(đổi biến x t2tan )
I I I1 2
34 6ln2
2
Vậy: I I I1 2
34 6ln2
2
8 ln
13
xI dx
xCâu 177.
8 ln
13
xI dx
x
u x dxdu
dx xdvv xx
ln
2 11
xI x x dx
x
88
33
12 1ln 2
Đặt
u x dxdu
dx xdv v xx
ln
2 11
xI x x dx
x
88
33
12 1ln 2
xJ dx
x
8
3
1 + Tính
xJ dx
x
8
3
1 t x 1 . Đặt t x 1
t dtJ dt
t t
3 32
2 22 2
2 12 1 2 ln3 ln2
1 1
I 6ln8 4ln3 2(2 ln3 ln2) 20ln2 6ln3 4
t dt
J dtt t
3 32
2 22 2
2 12 1 2 ln3 ln2
1 1
I 6ln8 4ln3 2(2 ln3 ln2) 20ln2 6ln3 4
dxxx
xI
2
1
3
2
ln1
Câu 178. dxxx
xI
2
1
3
2
ln1
I xdxxx
2
31
1 1ln
Ta có: I xdx
xx
2
31
1 1ln
u x
dv dxxx3
ln
1 1( )
. Đặt
u x
dv dxxx3
ln
1 1( )
I x x xdxxx x
22
4 511
1 1 1ln ln ln
4 4
I x x x dxxx x
22
4 511
1 1 1ln ln ln
4 4
21 63 1
ln2 ln 264 4 2
= 21 63 1ln2 ln 2
64 4 2
exx x x
I e dxx
2
1
ln 1 Câu 179.
e xx x x
I e dxx
2
1
ln 1
e e e xx x e
I xe dx e xdx dx H K Jx
1 1 1
ln Ta có: e e e x
x x eI xe dx e xdx dx H K J
x1 1 1
ln
e ex x e x eH xe dx xe e dx e e1
1 1
( 1) + e e
x x e x eH xe dx xe e dx e e11 1
( 1)
e e ex xex x e ee e
K e xdx e x dx e dx e Jx x1
1 1 1
ln ln + e e ex xe
x x e ee eK e xdx e x dx e dx e J
x x11 1 1
ln ln
e e e eI H K J e e e J J e1 1 Vậy: e e e eI H K J e e e J J e1 1 .
http://megabook.vn/
Trang 41
x xI dx
x
2
3
4
cos
sin
Câu 180.x x
I dxx
2
3
4
cos
sin
x
x x2 3
1 2cos
sin sin
u x
xdv dx
x3
cos
sin
. Đặt
u x
xdv dx
x3
cos
sin
du dx
vx2
1
2sin
du dx
vx2
1
2sin
xx
2
2
4
1 1.
2 sin
I = x
x
2
2
4
1 1.
2 sin
dxx
x
2 2
2
44
1 1 1( ) cot
2 2 2 2 2sin
+
dx x
x
2 2
2
44
1 1 1( ) cot
2 2 2 2 2sin
1
2=
1
2.
x xI dx
x
4
30
sin
cos
Câu 181.x x
I dxx
4
30
sin
cos
u x du dx
xdv dx v
x x3 2
sin 1
cos 2.cos
x dxI x
x x
44 4
2 200 0
1 1 1tan
2 4 2 4 22cos cos
Đặt:
u x du dx
xdv dx v
3 x x2
sin 1
cos 2.cos
x dxI x
x x
44 4
2 200 0
1 1 1tan
2 4 2 4 22cos cos
dxx
xxI
2
0
2
2sin1
)sin(
Câu 182. dxx
xxI
2
0
2
2sin1
)sin(
x xI dx dx H K
x x
22 2
0 0
sin
1 sin2 1 sin2
Ta có: x x
I dx dx H Kx x
22 2
0 0
sin
1 sin2 1 sin2
x xH dx dx
xx
2 2
20 01 sin2
2cos4
+ x x
H dx dxx
x
2 2
20 01 sin2
2cos4
u xdu dx
dxdv
v xx2
1tan
2cos 2 44
. Đặt:
u x du dx
dxdv
v xx2
1tan
2cos 2 44
xH x x
22
0 0
1tan ln cos
2 4 2 4 4
xK dx
x
22
0
sin
1 sin2
+ x
K dxx
22
0
sin
1 sin2
t x2
. Đặt t x
2
xK dx
x
22
0
cos
1 sin2
x
K dxx
22
0
cos
1 sin2
dxK x
x
2 2
20 0
12 tan 1
2 42cos
4
K1
2
I H K1
4 2
Vậy, I H K
1
4 2
.
http://megabook.vn/
Trang 42
x x x x
I dxx
3
20
(cos cos sin )
1 cos
Câu 183.
x x x x
I dxx
3
20
(cos cos sin )
1 cos
x x x x xI x dx x x dx dx J K
x x
2
2 20 0 0
cos (1 cos ) sin .sin.cos .
1 cos 1 cos
Ta có:
x x x x xI x dx x x dx dx J K
x x
2
2 20 0 0
cos (1 cos ) sin .sin.cos .
1 cos 1 cos
J x x dx
0
.cos .
+ Tính J x x dx
0
.cos . u x
dv xdxcos
. Đặt
u x
dv xdxcos
J x x x dx x
0 00
( .sin ) sin . 0 cos 2
J x x x dx x
0 00
( .sin ) sin . 0 cos 2
x xK dx
x20
.sin
1 cos
+ Tính x x
K dxx2
0
.sin
1 cos
x t dx dt
t t t t x xK dt dt dx
t t x2 2 20 0 0
( ).sin( ) ( ).sin ( ).sin
1 cos ( ) 1 cos 1 cos
x x x x dx x dxK dx K
x x x2 2 20 0 0
( ).sin sin . sin .2
21 cos 1 cos 1 cos
. Đặt x t dx dt
t t t t x xK dt dt dx
t t x2 2 20 0 0
( ).sin( ) ( ).sin ( ).sin
1 cos ( ) 1 cos 1 cos
x x x x dx x dxK dx K
2 x x x2 20 0 0
( ).sin sin . sin .2
21 cos 1 cos 1 cos
t x dt x dxcos sin . dt
Kt
1
21
2 1
Đặt t x dt x dxcos sin . dt
Kt
1
21
2 1
t u dt u du2tan (1 tan )
u duK du u
u
2 24 44
2
44 4
(1 tan ).
2 2 2 41 tan
, đặt t u dt u du2tan (1 tan )
u duK du u
u
2 24 44
2
44 4
(1 tan ).
2 2 2 41 tan
I2
24
Vậy I
2
24
x x x xI dx
x x
2
32
3
( sin )sin
(1 sin )sin
Câu 184.
x x x xI dx
x x
2
32
3
( sin )sin
(1 sin )sin
x x x x dxI dx dx H K
xx x x
2 2 223 3 3
2 2
3 3 3
(1 sin ) sin
1 sin(1 sin )sin sin
Ta có:
x x x x dxI dx dx H K
xx x x
2 2 223 3 3
2 2
3 3 3
(1 sin ) sin
1 sin(1 sin )sin sin
xH dx
x
2
32
3sin
+
xH dx
x
2
32
3sin
u xdu dx
dxdv v x
x2cot
sin
. Đặt
u x du dx
dxdv v x
x2cot
sin
H3
H
3
dx dx dxK
x xx
2 2 2
3 3 3
23 3 3
3 21 sin
1 cos 2cos2 4 2
+ dx dx dx
K x x
x
2 2 2
3 3 3
23 3 3
3 21 sin
1 cos 2cos2 4 2
I 3 23
Vậy I 3 2
3
x xI dx
x
23
0
sin
1 cos2
Câu 185.x x
I dxx
23
0
sin
1 cos2
http://megabook.vn/
Trang 43
x x x xI dx dx dx H K
x x x
2 23 3 3
0 0 2 0 2
sin sin
1 cos2 2cos 2cos
Ta có: x x x x
I dx dx dx H Kx x x
2 23 3 3
0 0 2 0 2
sin sin
1 cos2 2cos 2cos
x xH dx dx
x x
3 30 2 0 2
1
22cos cos
+ x x
H dx dxx x
3 30 2 0 2
1
22cos cos
u x
du dxdx
dv v xx2
tancos
. Đặt
u x du dx
dxdv v x
x2tan
cos
H x x xdx x3 3300 0
1 1 1tan tan lncos ln2
2 2 223 23
xK dx xdx
x
223 3
0 2 0
sin 1tan
22cos
x x 30
1 1tan 3
2 2 3
+ x
K dx xdxx
223 3
0 2 0
sin 1tan
22cos
x x 30
1 1tan 3
2 2 3
I H K
1 1 3 1 1ln2 3 ( 3 ln2)
2 2 3 6 22 3
Vậy:
I H K
1 1 3 1 1ln2 3 ( 3 ln2)
2 2 3 6 22 3
I x x dx3
0
1sin 1. Câu 186. I x x dx3
0
1sin 1.
t x 1 Đặt t x 1 I t t tdt t tdt x xdx2 2 2
2 2
1 1 1
.sin .2 2 sin 2 sin I t t tdt t tdt x xdx2 2 2
2 2
1 1 1
.sin .2 2 sin 2 sin
du xdxu xv xdv xdx
2 42cossin
Đặt du xdxu xv xdv xdx
2 42cossin
I x x x xdx22
2
11
2 cos 4 cos I x x x xdx22
2
11
2 cos 4 cos
u x du dx
dv xdx v x
4 4
cos sin
Đặt
u x du dx
dv xdx v x
4 4
cos sin
. Từ đó suy ra kết quả.
xxI e dx
x
2
0
1 sin.
1 cos
Câu 187.
xxI e dx
x
2
0
1 sin.
1 cos
xxe dx x
I e dxx x
2 2
20 0
1 sin
2 1 coscos
2
e x
xdx xI e dx
x x
2 2
20 0
1 sin
2 1 coscos
2
x x
x xx
I e dx e dxxx
2 2
120 0
2sin .cossin 2 2
1 cos2cos
2
xx
e dx2
0
tan2
+ Tính x x
x xx
I e dx e dxxx
2 2
120 0
2sin .cossin 2 2
1 cos2cos
2
xx
e dx2
0
tan2
xe dxI
x
2
220
1
2cos
2
+ Tính xe dx
I x
2
220
1
2cos
2
xxu e
du e dxdx
dv xvx2 tan
2cos 22
. Đặt
u x xe
du e dxdx
dv xvx2 tan
2cos 22
xxI e edx
22
20
tan2
xxI e e dx
22
20
tan2
I I I e21 2
Do đó: I I I e21 2
.
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Trang 44
x
xI dx
e x
20
cos
(1 sin2 )
Câu 188.x
xI dx
e x
20
cos
(1 sin2 )
x
xI dx
e x x
20 2
cos
(sin cos )
x
xI dx
e x x
20 2
cos
(sin cos )
x x
x x x dxu due e
dx xdv vx xx x 2
cos (sin cos )
sin
sin cos(sin cos )
x x x
x x xdx xdxI
x xe e e
2 22
0 0 0
cos sin sin sin.sin cos
. Đặt x x
x x x dxu due e
dx xdv v x xx x 2
cos (sin cos )
sin
sin cos(sin cos )
x x x
x x xdx xdxI
x xe e e
2 22
0 0 0
cos sin sin sin.sin cos
x x
u x du xdx
dxdv v
e e
1 1
1 1
sin cos
1
Đặt
x x
u x du xdx
dxdv v
e e
1 1
1 1
sin cos
1
x x x
xdx xdxI x
e e ee
2 22
0 0 02
1 cos 1 cossin .
x x x
xdx xdxI x
e e ee
2 22
0 0 02
1 cos 1 cossin .
x x
u x du xdx
dxdv v
e e
2 2
1 1
cos sin
1
x x
xdxI x I I e
e ee e
222
0 02 2
1 1 sin 1cos . 1 2 1
e
I2 1
2 2
Đặt
x x
u x du xdx
dxdv v
e e
2 2
1 1
cos sin
1
x x
xdxI x I I e
e ee e
222
0 02 2
1 1 sin 1cos . 1 2 1
e
I2 1
2 2
x
x xI dx
6 64
4
sin cos
6 1
Câu 189.
x
x xI dx
6 64
4
sin cos
6 1
t x Đặt t x dt dx dt dx t x
t x
t t x xI dt dx
6 6 6 64 4
4 4
sin cos sin cos6 6
6 1 6 1
t x
t x
t t x xI dt dx
6 6 6 64 4
4 4
sin cos sin cos6 6
6 1 6 1
x
x
x xI dx x x dx
6 64 46 6
4 4
sin cos2 (6 1) (sin cos )
6 1
x dx
4
4
5 3cos4
8 8
5
16
x
x
x xI dx x x dx
6 64 46 6
4 4
sin cos2 (6 1) (sin cos )
6 1
x dx
4
4
5 3cos4
8 8
5
16
I5
32
.
x
xdxI
46
6
sin
2 1
Câu 190.x
xdxI
46
6
sin
2 1
x x x
x x x
xdx xdx xdxI I I
04 4 46 6
1 20
6 6
2 sin 2 sin 2 sin
2 1 2 1 2 1
Ta có: x x x
x x x
xdx xdx xdxI I I
04 4 46 6
1 20
6 6
2 sin 2 sin 2 sin
2 1 2 1 2 1
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Trang 45
x
x
xdxI
0 4
1
6
2 sin
2 1
+ Tính x
x
xdxI
0 4
1
6
2 sin
2 1
x t t
t t x
t t xI dt dt dx
0 0 04 4 4
1
6 6 6
2 sin ( ) sin sin
2 1 2 1 2 1
x
x x
xdx xdxI xdx x dx
4 46 6 6 64 2
0 0 0 0
sin 2 sin 1sin (1 cos2 )
42 1 2 1
x x dx6
0
1(3 4cos2 cos4 )
8
4 7 3
64
. Đặt x t t
t t x
t t xI dt dt dx
0 0 04 4 4
1
6 6 6
2 sin ( ) sin sin
2 1 2 1 2 1
x
x x
xdx xdxI xdx x dx
4 46 6 6 64 2
0 0 0 0
sin 2 sin 1sin (1 cos2 )
42 1 2 1
x x dx6
0
1(3 4cos2 cos4 )
8
4 7 3
64
e
I x dx
1
cos(ln )
Câu 191.
e
I x dx
1
cos(ln )
t tt x x e dx e dtln Đặt t tt x x e dx e dtln
tI e tdt
0
cos
tI e tdt
0
cos e1
( 1)2
= e1
( 1)2
(dùng pp tích phân từng phần).
xI e x xdx22
sin 3
0
.sin .cos
Câu 192.xI e x xdx
22sin 3
0
.sin .cos
t x2sin Đặt t x2sintI e t dt e
1
0
1 1(1 )
2 2
tI e t dt e1
0
1 1(1 )
2 2 (dùng tích phân từng phần)
I x dx4
0
ln(1 tan )
Câu 193. I x dx4
0
ln(1 tan )
t x4
Đặt t x
4
I t dt
4
0
ln 1 tan4
I t dt4
0
ln 1 tan4
tdt
t
4
0
1 tanln 1
1 tan
=
t dt
t
4
0
1 tanln 1
1 tan
dt
t
4
0
2ln
1 tan
= dt
t
4
0
2ln
1 tan
dt t dt4 4
0 0
ln2 ln(1 tan )
= dt t dt4 4
0 0
ln2 ln(1 tan )
t I40.ln2
I2 ln24
= t I40.ln2
I2 ln24
I ln2
8
I ln2
8
.
I x x dx2
0
sin ln(1 sin )
Câu 194. I x x dx2
0
sin ln(1 sin )
xu x du dx
xdv xdxv x
1 cosln(1 sin )
1 sinsincos
Đặt x
u x du dxxdv xdx
v x
1 cosln(1 sin )
1 sinsincos
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Trang 46
x xI x x x dx dx x dx
x x
22 2 2
0 0 0
cos 1 sincos .ln(1 sin ) cos . 0 (1 sin ) 12
1 sin 1 sin 20
x xI x x x dx dx x dx
x x
22 2 2
0 0 0
cos 1 sincos .ln(1 sin ) cos . 0 (1 sin ) 12
1 sin 1 sin 20
x xI dx
x
4
0
tan .ln(cos )
cos
Câu 195.x x
I dxx
4
0
tan .ln(cos )
cos
t xcos Đặt t xcos dt xdxsin dt xdxsin t t
I dt dtt t
1
12
2 211
2
ln ln
t tI dt dt
t t
1
12
2 211
2
ln ln .
u t
dv dtt2
ln
1
Đặt
u t
dv dtt2
ln
1
du dtt
vt
1
1
du dt
t
v t
1
1
I2
2 1 ln22
I 2
2 1 ln22
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Trang 47
f x f x x4( ) ( ) cos
TP6: TÍCH PHÂN HÀM SỐ ĐẶC BIỆT
Câu 196. Cho hàm số f(x) liên tục trên R và f x f x x4( ) ( ) cos với mọi xR.
I f x dx2
2
( )
Tính: I f x dx2
2
( )
.
f x dx f t dt f t dt f x dx2 2 2 2
2 2 2 2
( ) ( )( ) ( ) ( )
Đặt x = –t 2
f x dx f t dt f t dt f x dx2 2 2
2 2 2 2
( ) ( )( ) ( ) ( )
f x dx f x f x dx xdx2 2 2
4
2 2 2
2 ( ) ( ) ( ) cos
2
f x dx f x f x dx xdx2 2
4
2 2 2
2 ( ) ( ) ( ) cos
I3
16
I
3
16
x x x4 3 1 1cos cos2 cos4
8 2 8 Chú ý: 4 x x x
3 1 1cos cos2 cos4
8 2 8 .
f x f x x( ) ( ) 2 2cos2 Câu 197. Cho hàm số f(x) liên tục trên R và f x f x x( ) ( ) 2 2cos2 , với mọi xR.
I f x dx
3
2
3
2
( )
Tính: I f x dx
3
2
3
2
( )
.
I f x dx f x dx f x dx
3 302 2
03 3
2 2
( ) ( ) ( )
Ta có : I f x dx f x dx f x dx
3 302 2
03 3
2 2
( ) ( ) ( )
(1)
I f x dx0
1
32
( )
+ Tính : I f x dx0
1
32
( )
x t dx dt . Đặt x t dx dt I f t dt f x dx
3 32 2
10 0
( ) ( )
I f t dt f x dx
3 32 2
10 0
( ) ( )
I f x f x dx x x dx
3 3 32 2 2
0 0 0
( ) ( ) 2 1 cos2 2 cos
xdx xdx
32 2
0
2
2 cos cos
x x20
3
22 sin sin 6
2
Thay vào (1) ta được: I f x f x dx x x dx
3 3 32 2 2
0 0 0
( ) ( ) 2 1 cos2 2 cos
xdx xdx
32 2
0
2
2 cos cos
x x20
3
22 sin sin 6
2
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Trang 48
xI dx
x x
4
2
4
sin
1
Câu 198.
xI dx
x x
4
2
4
sin
1
I x xdx x xdx I I4 4
21 2
4 4
1 sin sin
I x xdx x xdx I I4 4
21 2
4 4
1 sin sin
I x xdx4
21
4
1 sin
+ Tính I x xdx4
21
4
1 sin
I1 0. Sử dụng cách tính tích phân của hàm số lẻ, ta tính được I1 0 .
I x xdx4
2
4
sin
+ Tính I x xdx4
2
4
sin
I2
22
4 . Dùng pp tích phân từng phần, ta tính được: I2
22
4
I2
24 Suy ra: I
22
4 .
5
2
3 2 1
1 1
x
x
e x xI dx
e x x
Câu 199.
5
2
3 2 1
1 1
x
x
e x xI dx
e x x
5 5 5 5
2 2 2 2
3 2 1 1 1 2 1 2 1
1 1 1 1 1 1
x x x x
x x x
e x x e x x e x e xI dx dx dx dx
e x x e x x e x x
5 5
2 2
5 2 1 2 13
2 1( 1 1) 1( 1 1)
x x
x x
e x e xx dx dx
x e x x e x
5 5 5 5
2 2 2 2
3 2 1 1 1 2 1 2 1
1 1 1 1 1 1
x x x x
x x x
e x x e x x e x e xI dx dx dx dx
e x x e x x e x x
5 5
2 2
5 2 1 2 13
2 1( 1 1) 1( 1 1)
x x
x x
e x e xx dx dx
x e x x e x
2 11 1
2 1
x
x e xt e x dt dx
x5
2
52 1 5
22
1
2 12 2 13 3 2ln 3 2ln
11
e
e
e eI dt I t
t ee
Đặt 2 1
1 1 2 1
x
x e xt e x dt dx
x5
2
52 1 5
22
1
2 12 2 13 3 2ln 3 2ln
11
e
e
e eI dt I t
t ee
xI dx
x x x
24
20 ( sin cos )
Câu 200.x
I dxx x x
24
20 ( sin cos )
.
x x xI dx
x x x x
4
20
cos.
cos ( sin cos )
x x x
I dxx x x x
4
20
cos.
cos ( sin cos )
xu
xx x
dv dxx x x 2
coscos
( sin cos )
. Đặt
xu
xx x
dv dxx x x 2
coscos
( sin cos )
x x xdu dx
x
vx x x
2
cos sin
cos1
sin cos
x x xdu dx
x
v x x x
2
cos sin
cos1
sin cos
x dxI dx
x x x x x
44
20 0
cos ( sin cos ) cos
x dx
I dxx x x x x
44
20 0
cos ( sin cos ) cos
4
4
=
4
4
.
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