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Tuyển tập 100 hệ phương trình thường gặp (2015-2016) - Megabook.vn
200 Bài tập Tích phân - Megabook.vn
1. Trang 1
TP1: TÍCH PHÂN HÀM SỐ HỮU TỈ
x
I dx
x x
2 2
2
1 7 12
Dạng 1: Tách phân thức
xxx
ddd
x x
222 222
2
111 777 222
I dx
x x
2
1
16 9
1
4 3
Câu 1. III xxx
x x2
111
ddd
x x
222
1
x x x
2
116ln 4 9ln 3 III xxx
x x1
111666 999
111
444 333
xxx xxx 111111666 nnn 444 999lllnnn 333 1 25ln2 16ln3 = xxx
222
lll 111 = 222555lllnnn222 111666lllnnn333
dx
I
x x
2
5 3
1
.
x x
222
5 3
1
x
xx x x x3 2 3 2
1 1 1
( 1) 1
Câu 2.
dddxxx
III
x x5 3
1
xx x x x3 2 3 2
111 111
( 1) 1
I x x
x
2
2
21 1 3 1 3
ln ln( 1) ln2 ln5
2 2 2 812
Ta có:
xxx
xx x x x3 2 3 2
111
( 1) 1
III xxx xxx
x2
222111 111 333 111 333
lllnnn lllnnn((( 111))) lllnnn222 lllnnn555
2 2 2 812
x
I dx
x x x
5 2
3 2
4
3 1
2 5 6
x
222
2 2 2 2 812
xxx
III dddxxx
x x x
555 222
3 2
4
111
2 5 6
I
2 4 13 7 14
ln ln ln2
3 3 15 6 5
Câu 3.
x x x3 2
4
333
2 5 6
444 111 777 111
lllnnn lllnnn222
3 3 15 6 5
xdx
I
x
1
0 3
( 1)
III
222 333 444
lllnnn
3 3 15 6 5
ddd
x
111
0 3
( 1)
x x
x 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
III
x0 3
( 1)
xxx xxx
xxx xxx
x x
222 333
3 3
((( 111))) ((( 111)))
( 1) ( 1)
ddd
0
111 Ta có:
x x3 3
111 111
( 1) ( 1)
III xxx xxx xxx
111 222 333
0
111
((( ))) ((( 111)))
888
x
I dx
x
2
4
( 1)
(2 1)
Dạng 2: Đổi biến số
III ddd
x
222
4
((( 111
(2 1)
x x
f x
x x
2
1 1 1
( ) . .
3 2 1 2 1
Câu 5.
xxx
xxx
x 4
)))
(2 1)
222
111 111
3 2 1 2 1
x
I C
x
3
1 1
9 2 1
Ta có:
xxx xxx
fff xxx
xxx xxx
111
((( ))) ... ...
3 2 1 2 1
xxx
III CCC
x
333
111 111
9 2 1
x
I dx
x
991
101
0
7 1
2 1
x9 2 1
xxx
III dddxxx
x
101
0
777 111
2 1
x dx x x
I d
x x xx
99 991 1
2
0 0
7 1 1 7 1 7 1
2 1 9 2 1 2 12 1
Câu 6.
x
999999111
101
0 2 1
dddxxx xxx xxx
ddd
x x xx
999
2
0 0
2 1 9 2 1 2 12 1
x
x
100
1001 1 7 1 11
2 1
09 100 2 1 900
x
I dx
x
1
2 2
0
5
( 4)
xxx
III
x x xx
999999 999111 111
2
0 0
777 111 111 777 111 777 111
2 1 9 2 1 2 12 1
x
x
100
1001 1 7 1 11
2 1
09 100 2 1 900
xxx
III dddxxx
x
111
2 2
0
555
( 4)
Câu 7.
x2 2
0 ( 4)
t x2
4 I
1
8
Đặt ttt xxx222
444
111
888
III
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2. x
I dx
x
1 7
2 5
0 (1 )
Trang 2
III ddd
x
111 777
2 5
000 (((
t x dt xdx2
1 2 Câu 8.
xxx
xxx
x2 5
111 )))
dddttt222
111
t
I dt
t
2 3
5 5
1
1 ( 1) 1 1
.
2 4 2
Đặt ttt xxx xxxdddxxx222 III dddttt
t
222 333
5 5
1
111 ((( 111))) 111 111
2 4 2
I x x dx
1
5 3 6
0
(1 )
ttt
t5 5
1
...
2 4 2
ddd
111
0
(((
dt t t
t x dt x dx dx I t t dt
x
1 7 8
3 2 6
2
0
1 1 1
1 3 (1 )
3 3 7 8 1683
Câu 9. III xxx xxx xxx555 333 666
0
111 )))
ttt ttt ttt
ttt ddd
x
111 777 888
333 222 666
2
0
111 111
111 333 111 )))
3 3 7 8 1683
I dx
x x
4
3
4
1
1
( 1)
Đặt
ddd
ttt xxx dddttt xxx dddxxx dddxxx III ttt ttt
x2
0
111
(((
3 3 7 8 1683
III dddxxx
x x
444
4
1
111
( 1)
t x2
Câu 10.
x x
333
4
1 ( 1)
222 t
I dt
t t
3
2
1
1 1 1 3
ln
2 4 21
Đặt ttt xxx
ttt
III ddd
t t2
1
111
lllnnn
2 4 21
dx
I
x x
2
10 2
1 .( 1)
ttt
t t
333
2
1
111 111 333
2 4 21
xxx
III
x x10 2
...((( 111)))
x dx
I
x x
2 4
5 10 2
1
.
.( 1)
Câu 11.
ddd
x x
222
10 2
111
ddd
x x
444
5 10 2
1 .( 1)
t x5
xxx xxx
III
x x
222
5 10 2
1
...
.( 1)
555 dt
I
t t
32
2 2
1
1
5 ( 1)
. Đặt ttt xxx
ttt
III
t t
333222
2 2
1
111
555 ((( 111)))
x
I dx
x x
2 7
7
1
1
(1 )
ddd
t t2 2
1
xxx
ddd
x x
222 777
7
1 (1 )
x x
I dx
x x
2 7 6
7 7
1
(1 ).
.(1 )
Câu 12. III xxx
x x7
1
111
(1 )
xxx xxx
x x
222 777 666
7 7
1
111
.(1 )
t x7
III dddxxx
x x7 7
1
((( )))...
.(1 )
777 t
I dt
t t
128
1
1 1
7 (1 )
. Đặt ttt xxx
ttt
III dddttt
t t
111 888
1
111 111
7 (1 )
dx
I
x x
3
6 2
1 (1 )
t t
222
1
7 (1 )
dddxxx
III
x x
333
6 2
111 )))
x
t
1
Câu 13.
x x6 2
111 (((
111 t
I dt t t dt
t t
3
163
4 2
2 2
1 3
3
1
1
1 1
Đặt : xxx
ttt
III ddd dddttt
t t
333
111666333
444 222
2 2
1 3
3
111
111
1 1
117 41 3
135 12
ttt
ttt ttt ttt
t t2 2
1 3
3
1 1
111 444
135 12
x
I dx
x
2 2001
2 1002
1
.
(1 )
=
111 777 111 333
135 12
xxx
dddxxx
x
222 000000111
2 1002
1 (1 )
x
I dx dx
x x
x
x
2 22004
3 2 1002 1002
1 1 3
2
1
. .
(1 ) 1
1
Câu 14. III
x
222
2 1002
1
...
(1 )
xxx
III ddd xxx
x x
x
x
222 222444
3 2 1002 1002
1 1 3
2
... ...
(1 ) 1
1
t dt dx
x x2 3
1 2
1 xxx ddd
x x
x
x
222000000
3 2 1002 1002
1 1 3
2
111
(1 ) 1
1
ttt dddttt dddxxx
x x2 3
111 222
. Đặt
x x2 3
111
x xdx
I
x x
1 2000
2 2000 2 2
0
1 .2
2 (1 ) (1 )
.
xxx xxxdddxxx
III
x x
111 000
2 2000 2 2
0
111
222 (((111 ))) (((111 )))
Cách 2: Ta có:
x x
222000 000
2 2000 2 2
0
...222
t x dt xdx2
1 2
t
I dt d
t tt t
10002 21000
1000 2 1001
1 1
1 ( 1) 1 1 1 1
1 1
2 2 2002.2
. Đặt ttt xxx dddttt xxxdddxxx222
111 222
ttt
III ddd
t tt t
000000
1000 2 1001
1 1
((( ))) 111 111 111
111 111
2 2 2002.2
x
I dx
x
2 2
4
1
1
1
ttt ddd
t tt t
111 000000222 222111 000000
1000 2 1001
1 1
111 111 111
2 2 2002.2
xxx
III dddxxx
x
222
4
1
111
1
Câu 15.
x
222
4
11
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3. x x
x x
x
2 2
4
2
2
1
1
1
11
Trang 3
xxx xxx
x x
x
222
4
2
2
111
111
111
11
t x dt dx
x x2
1 1
1
Ta có:
x x
x
222
4
2
2
11
t x dt dx
222
1 1
1
dt
I dt
t tt
3 3
2 2
2
1 1
1 1 1
2 2 2 22
t
t
3
1 2 1 2 1
.ln ln2
2 2 2 2 2 2 11
. Đặt t x dt dx
xxx xxx
1 1
1
dt
I dt
3 3
2 2
111 111
1 1 1
t
3
1 2 1 2 1
.ln ln2
222 222 111
x
I dx
x
2 2
4
1
1
1
dt
I dt
ttt tttttt
3 3
2 2
222
1 1 1
222 222 222 222222
t
ttt
3
1 2 1 2 1
.ln ln2
222 222 222 222111
x
I dx
2 2
1
111
x x
x x
x
2 2
4
2
2
1
1
1
11
Câu 16.
x
I dx
xxx
2 2
444
111
1
xxx xxx
x x
x
222
4
2
2
111
111
111
11
t x dt dx
x x2
1 1
1
Ta có:
x x
x
222
4
2
2
11
t x dt dx
xxx 222
1 1
1
dt
I
t
5
2
2
2 2
. Đặt t x dt dx
xxx
1 1
1
dt
I
5
2
2 2
dt
I
ttt
5
2
222
2 2
du
t u dt
u2
2tan 2
cos
.
uuu
ttt uuu dddttt
u2
222 aaannn 222
cos
u u u u1 2
5 5
tan 2 arctan2; tan arctan
2 2
Đặt
ddd
u2
ttt
cos
111 222aaarrr aaa ttt aaarrr
2 2
u
u
I d u u u
2
1
2 1
2 2 2 5
( ) arctan arctan2
2 2 2 2
; uuu uuu uuu uuu
555 555
tttaaannn 222 cccttt nnn222;;; aaannn ccctttaaannn
2 2
u
222
ddduuu uuu uuu
1
222 222 222 555
aaarrrccc aaarrrccc
2 2 2 2
x
I dx
x x
2 2
3
1
1
uuu
u
III
1
222 111((( ))) tttaaannn tttaaannn222
2 2 2 2
ddd
x x3
1
xI dx
x
x
2 2
1
1
1
1
Câu 17.
xxx
III xxx
x x
222 222
3
1
111
dddxxx
x
x
222 222
1
1
t x
x
1
Ta có: xxxIII
x
x
1
111
111
1
xxx
xxx
111
. Đặt ttt I
4
ln
5
x
I dx
x
1 4
6
0
1
1
III
444
lllnnn
555
xxx
III ddd
x
111
6
0
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. xxx
x
444
6
0 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 x
I dx dx
x x
1 1 3
2 3 2
0 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 6
1 1 ( 1)( 1) 1 1 1
d x
I dx dx
1 1 3
1 1 ( ) 1
.
x
I dx
x
3
23
4
0 1
d x
I dx dx
xxx xxx
1 1 3
222 333 222
000 000
1 1 ( ) 1
.
333 444 333 444 333111 ((( ))) 111
xxx
III dddxxx
x
333
222333
4
000 111
x
I dx dx
x x x x
3 3
23 3
2 2 2 2
0 0
1 1 1 1
ln(2 3)
2 4 12( 1)( 1) 1 1
Câu 19.
x4
xxx dddxxx
x x x x
333 333
222333 333
2 2 2 2
0 0
111
lllnnn(((222
2 4 12( 1)( 1) 1 1
xdx
I
x x
1
4 2
0 1
xxx
III ddd
x x x x2 2 2 2
0 0
111 111 111
333)))
2 4 12( 1)( 1) 1 1
dddxxx
x x
111
4 2
000
Câu 20.
xxx
III
x x4 2
111
t x2
dt dt
I
t t
t
1 1
2 22
0 0
1 1
2 2 6 31 1 3
2 2
. Đặt ttt xxx222
ttt dddttt
III
t t
t
111 111
2 22
0 0
111 111
2 2 6 31 1 3
2 2
ddd
t t
t
2 22
0 0
2 2 6 31 1 3
2 2
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4. x
I dx
x x
1 5
22
4 2
1
1
1
Trang 4
dddxxx
x x
222
4 2
1 1
x x
x x x
x
2 2
4 2
2
2
1
1
1
11 1
Câu 21.
xxx
III
x x
111 555
222
4 2
1
111
1
x x x
x
222 222
4 2
2
2
111
111
11 1
t x dt dx
x x2
1 1
1
Ta có:
xxx xxx
x x x
x
4 2
2
2
111
11 1
t x dt dx
222
1 1
1
dt
I
t
1
2
0 1
. Đặt t x dt dx
xxx xxx
1 1
1
dt
I
1
222
0 1
du
t u dt
u2
tan
cos
dt
I
ttt
1
0 1
ttt uuu ttt
u2
ttt
cos
I du
4
0
4
. Đặt
ddduuu
ddd
u2
aaannn
cos
ddd
0
4
III uuu
444
0
4
TP2: TÍCH PHÂN HÀM SỐ VÔ TỈ
x
I dx
x x2
3 9 1
Dạng 1: Đổi biến số dạng 1
xxx
III ddd
x x2
3 9 1
x
I 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 x2
3 9 1
xxx
III 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 3
1 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 dx2
2 9 1 x d x x C
3
2 2 2 2
2
1 1
9 1 (9 1) (9 1)
18 27
+ xxx ddd CCC222 333
III xxx xxx xxx222 xxx ddd xxx xxx CCC
111 111
999 111 (((999 ))) 999 111)))
18 27
I x x C
3
2 321
(9 1)
27
+ ddd222
999 111
333
222 222 222 222
222111 (((
18 27
III xxx
111
999 111)))
27
x x
I dx
x x
2
1
xxx CCC
333
222 333222(((
27
xxx
III dddxxx
x x1
x x
dx
x x
2
1
x x
dx dx
x x x x
2
1 1
Câu 23.
xxx
x x
222
1
xxx
xxx
x x1
xxx xxx
dddxxx xxx
x x x x1 1
xxx
ddd
x x
222
1
ddd
x x x x
222
1 1
x
I dx
x x
2
1
1
.
xxx
III xxx
x x111
x x t x x2
1 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 dddttt
444
((( 111)))
3
ddd ttt
3 9 3 x x x x C
3
1
4 4
1 1
9 3
ttt ttt ttt CCC222 333444 444 444
((( 111)))
3 9 3
xxx xxx xxx CCC
333
111
9 3
x
I d x
x x
2
1
= xxx
444 444
111 111
9 3
xxx
III dddxxx
x x111
d x x
x x
2 (1 )
3 1
+
x x
222
ddd
x x
222 (((111 )))
3 1
x x C2
4
1
3
=
xxx xxx
x x3 1
222
3
I x x C
3
4
1
9
= xxx xxx CCC
444
111
3
I x x C
3
4
1
x
I dx
x
4
0
2 1
1 2 1
Vậy: I x x C
3
4
1
999
xxx
III dddxxx
x
444
01 2 1
t x2 1 Câu 24.
x0
222 111
1 2 1
ttt xxx
t
dt
t
3 2
1
2 ln2
1
Đặt 222 111
ttt
dddttt
t
333 222
1
222 lllnnn222
1
. I =
t1
1
.
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5. dx
I
x x
6
2 2 1 4 1
Trang 5
III
x x
666
222
t x4 1 Câu 25.
dddxxx
x x
222 111 444 111
Đặt ttt xxx444 111 I
3 1
ln
2 12
I x x dx
1
3 2
0
1
. III
333 111
lllnnn
222 111222
ddd
111
333 222
0
t x2
1 Câu 26. III xxx xxx xxx
0
111 ttt 111 I t t dt
1
2 4
0
2
15
Đặt: xxx222
ttt ddd
111
222 444
0
222
15
III ttt ttt
0
15
x
I dx
x
1
0
1
1
.
xxx
III dddxxx
x
111
0
111
1
t x
Câu 27.
x01
xxx dx t dt2 . Đặt ttt ddd ddd...
t t
dt
t
1 3
0
2
1
xxx ttt ttt222
t
111 333
0
t t dt
t
1
2
0
2
2 2
1
. I =
ttt ttt
dddttt
t0
222
111
ddd
t
111
222
0
11
4ln2
3
= ttt ttt ttt
t0
222
222 222
111
3
=
111111
444lllnnn222
3
x
I dx
x x
3
0
3
3 1 3
.
ddd
x x
333
0 3 1 3
t x tdu dx1 2
Câu 28.
xxx
III xxx
x x0
333
3 1 3
ttt xxx tttddduuu xxx111 222
t t
I dt t dt dt
tt t
2 2 23
2
1 1 1
2 8 1
(2 6) 6
13 2
3
3 6ln
2
Đặt ddd
ttt ttt
III dddttt ttt dddttt dddttt
tt t
222 222 222
2
1 1 1
222 888 111
((( 666 666
1113 2
333
333 666lllnnn
222
I x x dx
0
3
1
. 1
tt t
333
2
1 1 1
222 )))
3 2
ddd
000
333
1
...
t t
t x t x dx t dt I t dt
1
1 7 4
3 2 33
00
9
1 1 3 3( 1) 3
7 4 28
Câu 29. III xxx xxx xxx
1
111
ttt ttt
ttt xxx ttt xxx dddxxx ttt dddttt III ttt ttt
00
999
111 111 333 333((( 111))) 333
777 444 888
x
I dx
x x
5 2
1
1
3 1
Đặt ddd
111
111 777 444
333 222 333333
00
222
xxx
III dddxxx
x x
555 222
1
111
3 1
tdt
t x dx
2
3 1
3
Câu 30.
x x1 3 1
ttt
xxx dddxxx
222
333 111
333
t
tdt
I
t
t
2
2
4
2
2
1
1
3 2
.
31
.
3
dt
t dt
t
4 4
2
2
2 2
2
( 1) 2
9 1
t
t t
t
3
4 4
2 1 1 100 9
ln ln .
9 3 1 27 52 2
Đặt
tttddd
ttt
ttt
tttdddttt
III
t
t
2
2
111
111
333 222
...
3331
.
3
ddd
t
444 444
222
2
2 2
222
)))
999
t
t t
t
3
4 4
2 1 1 100 9
ln ln .
9 3 1 27 52 2
x x
I dx
x
3 2
0
2 1
1
t
t
222
222
444
2
2 1
.
3
ttt
ttt dddttt
t2
2 2
((( 111 222
111
t
t t
t
3
4 4
2 1 1 100 9
ln ln .
9 3 1 27 52 2
x x
I dx
xxx
3 2
000
2 1
111
x t x t2
1 1
Câu 31.
x x
I dx
3 2
2 1
222
dx tdt2 Đặt xxx ttt xxx ttt111 111 dddxxx tttdddttt222
t t t
I tdt t t dt t
t
2
2 22 2 2 5
4 2 3
11 1
2( 1) ( 1) 1 4 54
2 2 (2 3 ) 2
5 5
ttt
III tttddd ttt ttt dddttt ttt
t
222
222 555
11 1
((( 111 555
222 222 222 333 ))) 222
5 5
ttt ttt
ttt
t
222 222222 222
444 222 333
11 1
222 ))) ((( 111))) 111 444 444
(((
5 5
http://megabook.vn/
6. x dx
I
x x
1 2
0
2
( 1) 1
Trang 6
xxx dddxxx
III
x x
111
0
222
( 1) 1
t x t x tdt dx2
1 1 2
t t
I tdt t dt t
t tt
222 22 2 3
3
11 1
( 1) 1 1 16 11 2
.2 2 2 2
3 3
Câu 32.
x x
222
0 ( 1) 1
ttt ddd222
t t
I tdt t dt t
t tt
222 22 2 3
3
11 1
( 1) 1 1 16 11 2
.2 2 2 2
3 3
x
I dx
x
4
2
0
1
1 1 2
Đặt ttt xxx ttt xxx dddttt xxx111 111 222
t t
I tdt t dt t
t tt
222 22 2 3
3
11 1
( 1) 1 1 16 11 2
.2 2 2 2
3 3
ddd
x
2
0 1 1 2
dx
t x dt dx t dt
x
1 1 2 ( 1)
1 2
Câu 33.
xxx
III xxx
x
444
2
0
111
1 1 2
ddd
ddd ttt
x
111 ((( 111)))
1 2
t t
x
2
2
2
Đặt
xxx
ttt xxx ttt dddxxx ttt ddd
x
111 222
1 2
xxx
222
222
2
t t t t t t
dt dt t dt
tt t t
4 4 42 3 2
2 2 2
2 2 2
1 ( 2 2)( 1) 1 3 4 2 1 4 2
3
2 2 2
và
ttt ttt
2
ttt ttt ttt ttt ttt ttt
ttt dddttt ttt dddttt
tt t t
222
2 2 2
2 2 2
111 ((( 222 222)))((( 111))) 111 333 444 222 111 444 222
333
2 2 2
t
t t
t
2
1 2
3 4ln
2 2
Ta có: I = ddd
tt t t
444 444 444333 222
2 2 2
2 2 2
2 2 2
ttt
222
333 lllnnn
2 2
=
ttt
ttt
ttt
111 222
444
2 2
1
2ln2
4
x
I dx
x
8
2
3
1
1
=
111
222lllnnn222
444
III dddxxx
x2
3 1
x
I dx
x x
8
2 2
3
1
1 1
Câu 34.
xxx
x
888
2
3
111
1
x x2 2
3 1 1
x x x
8
2 2
3
1 ln 1
xxx
III dddxxx
x x
888
2 2
3
111
1 1
888
222 222
3
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 dx
1
3 2
0
( 1) 2
= lll
III xxx xxx xxx dddxxx
111
333 222
0
I x x x dx x x x x x dx
1 1
3 2 2 2
0 0
( 1) 2 ( 2 1) 2 ( 1)
Câu 35.
0
((( 111))) 222
ddd ddd
111 111
333 222 222 222
0 0
((( ))) 222 ((( 222 ))) 222 ((( ))) t x x2
2 III xxx xxx xxx xxx xxx xxx xxx xxx xxx xxx
0 0
111 111 111 ttt xxx xxx222 I
2
15
. Đặt 222
15
III
222
15
x x x
I dx
x x
2 3 2
2
0
2 3
1
.
xxx xxx xxx
III dddxxx
x x2
0
222 333
1
x x x
I dx
x x
2 2
2
0
( )(2 1)
1
Câu 36.
x x
222 333 222
2
0 1
xxx xxx xxx
III xxx
x x
222
2
0
((( )))222 111)))
1
t x x2
1 I t dt
3
2
1
4
2 ( 1)
3
ddd
x x
222
2
0
(((
1
ttt xxx xxx 111 III ttt dddttt
1
444
222 ((( )))
333
. Đặt 222
333
222
1
111
x dx
I
x
2 3
3 2
0 4
.
ddd
x
3 2
0 4
t x x t xdx t dt
3 2 2 3 2
4 4 2 3
Câu 37.
xxx xxx
III
x
222 333
3 2
0 4
ttt xxx xxx ttt dddxxx ttt ttt444 444 222 333 I t t dt
3
2
4 3
4
3 3 8
( 4 ) 4 2
2 2 5
Đặt xxx ddd
333 222 222 333 222
III ddd
3
4
333
(((
2 2 5
dx
I
x x
1
2
11 1
ttt ttt ttt
3
222
444 333
4
333 888
444 ))) 444 222
2 2 5
ddd
x x
111
2
11 1
Câu 38.
xxx
III
x x2
11 1
http://megabook.vn/
7. x x x x
I dx dx
xx x
1 12 2
2 2
1 1
1 1 1 1
2(1 ) (1 )
x
dx dx
x x
1 1 2
1 1
1 1 1
1
2 2
I dx x x
x
1
1
1 1
1
1 1 1
1 ln | 1
2 2
x
I dx
x
1 2
2
1
1
2
Trang 7
xxx xxx xxx xxx
xxx xxx
222 222
111 111
111 111 111 111
(((
xxx
dddxxx xxx
111
111 111
111 111
111
xxx111
222 222
x
I dx
x
111 222
222
1
1
2
t x t x tdt xdx2 2 2
1 1 2 2
Ta có: d d
xx x
111 222
2 2
1 1
2(1 ) (1 )
ddd
x x
111 222
1 1
2 2
ddd
xxx
111
111
111
111
222 222
III
xxx
1 2
2
222
tttdddttt ddd111 111 222
t dt
t
2 2
2
2
0
2( 1)
+ xxx xxx xxx
111 111 111
nnn |||
xxx
dddxxx
111
111
ttt xxx ttt xxx xxx xxx222 222 222
222
2
t dt2
0
I 1
+ tttdddttt ddd111 111 222
ttt222
222 ((( )))
III
. Đặt ttt xxx ttt xxx xxx xxx222 222 222
222
dddttt
000
222 111
111
t x x2
1
I2=
ttt222 222
t x x2
1
Vậy: 111
xxx xxx222
111
x x
I dx
x
1
3 31
4
1
3
.
ttt xxx
xxx
III
xxx
1
3 31
I dx
x x
1
1
3
2 3
1
3
1 1
1 .
Cách 2: Đặt
444
111
333
III dddxxx
111
111
333
222 333
333
111 111
t
x2
1
1
.
xxx
dddxxx
111
111 ...
t
xxx
1
1 I 6Câu 39.
111
333 333111
III dddxxx
xxx xxx
111
111
333111 111
222
111
111 III Ta có: 111 ...
xxx
x
I dx
x
2 2
1
4
. Đặt ttt 666
x
I dxxx
x
2
1
4
x
I xdx
x
2 2
2
1
4
xxx
xxx
222
111
444
III xxxdddxxx
x t x tdt xdx2 2 2
4 4
.
III ddd
xxx
222 222
xxx ttt xxx dddttt xxxdddxxx444 444
t tdt t t
dt dt t
tt t t
00 0 02
2 2 2
33 3 3
( ) 4 2
(1 ) ln
24 4 4
Câu 40.
222
xxx
x xxx
222
111
444
ttt222 222 222
ttt
000000 2
222 222 222
333333 333
((( )))
( ))) lll
222444
2 3
3 ln
2 3
Ta có:
222 222
xxx ttt xxx dddttt xxxdddxxx444 444
ttt
ttt
000 000
333
444 222
444 444
222 333
333 lllnnn
222 333
x
I dx
x x
2 5
2 2
2 ( 1) 5
. Đặt t = ttt222 222 222
tttdddttt ttt ttt
dddttt dddttt ttt
ttt ttt
111 nnn
xxx
xxx x222 ((( )))
t x2
5
I =
222
(((
222 333
333 lllnnn
ddd ttt xxx 555
dt
I
t
5
2
3
1 15
ln
4 74
=
III xxx
222 555
111
222 dt
I
ttt
5
222
333
1 15
ln
444 777
Câu 41.
x
I dx
xxx
2 5
222 222
555
ttt xxx 555 III
444
x
I dx
x x
27
3 2
1
2
Đặt
222 ddd555
111
x
I dx
x x
2
3 2
1
2
ttt 111 555
lllnnn
xxx
7
222
t x6
t t
I dt dt
tt t t t
3 33
2 2 2
1 1
2 2 2 1
5 5 1
( 1) 1 1
2 5
5 3 1 ln
3 12
.
xxx
III dddxxx
222
333 222
111
ttt
ttt ttt
III dddttt ttt
t ttt222 222
111 111
222 222 222 111
555 555 111
) 111 111
222
lll
333 111
I dx
x x
1
2
0
1
1
Câu 42.
xxx
777
xxx666
ddd
tttttt ttt
333 333333
((( )))
555
555 333 111 nnn
222
III
xxx000
111
t x x x2
1
Đặt ttt
ttt ttt
III dddttt ttt
ttt222
222 222 222 111
555 555 111
111
222
lll
333 111
x x x
dt
I t
t
1 3
1 3
1
1
2 3 2 3
ln(2 1) ln
2 1 3
ddd
333 333333
555
555 333 111 nnn
222
dddxxx
xxx
111
111
x x222
dddttt
III ttt
ttt 1
111
222 333 222 333
lllnnn(((222 111))) lllnnn
222
x
I dx
x x
3 2
2 2
0 (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 dddttt
III ttt
111
222 333 222 333
lllnnn(((222 111))) lllnnn
xxx
III xxx
111
x t2 1 I t dt
t t
4
2
3
42 36 4
2 16 12 42ln
3
111 333
111 333
ddd
xxx
333
222
000 (((
x t2 1 t dt
ttt ttt
444
333
42 36 4
2 16 12 42llln
3
Câu 44.
xxx
III xxx
111 (((222 111 )))
III
222
Đặt xxx ttt222 111 ddd
333
111 111 444 ttt ttt
444222 666 444
222 666 222 222 nnn
333
http://megabook.vn/
8. x
I dx
x x x x
3 2
0 2( 1) 2 1 1
Trang 8
ddd
x x x x
333 222
000 111 111
t x 1
Câu 45.
xxx
III xxx
x x x x
222((( ))) 222 111
t t dt
I t dt
t t
2 22 2
2
2
1 1
2 ( 1)
2 ( 1)
( 1)
t
2
3
1
2 2
( 1)
3 3
Đặt ttt xxx 111
ddd
ttt dddttt
t t
222 222222 222
2
1 1
)))
111
( 1)
ttt
1
((( 111
3 3
x x x
I d x
x
32 2 3
4
1
2011
ttt ttt ttt
III
t t
222
2
1 1
222 ((( 111
222 ((( )))
( 1)
222
333
1
222 222
)))
3 3
xxx xxx xxx
III dddxxx
x
222 222
4
1
222000 111
xI dx dx M N
x x
3
2 2 2 22
3 3
1 1
1
1
2011
xM dx
x
3
2 2 2
3
1
1
1
Câu 46.
x
333 333
4
1
111
xxx ddd ddd
x x3 3
1 1
222000111
xM dx
x
3
2 2 2
3
1
1
1
t
x
3
2
1
1
Ta có: III xxx xxx MMM NNN
x x
333
222 222 222 222222
3 3
1 1
111
111
111
xM dx
x
3
2 2 2
3
1
1
1
t 3
2
1
1 M t dt
3
7
32
3
0
3 21 7
2 128
N dx x dx
x x
2 22 2 2 2
3
3 2
1 1 1
2011 2011 14077
2011
162
. Đặt t
xxx
3
2
1
1 ddd
777
333222
333
0
222
2 128
N dx x dx
x x
2 22 2 2 2
3
3 2
1 1 1
2011 2011 14077
2011
162
I
3
14077 21 7
16 128
MMM ttt ttt
333
0
333 111 777
2 128
N dx x dx
x x
2 22 2 2 2
3
3 2
1 1 1
2011 2011 14077
2011
162
I
3
14077 21 7
111 111222
I
3
14077 21 7
666 888
dx
I
x x
1
33 3
0 (1 ). 1
.
xxx
III
x x
33 3
0 (1 ). 1
t x
3 3
1
Câu 47.
ddd
x x
111
33 3
0 (1 ). 1
ttt 111
t dt
I dt
t t t t
3 3
2 22
2 2
1 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 4
1 1 1
3 342 3
33
1
1
11
1. 1
Đặt xxx
333 333 dddttt
dddttt
t t t t
222 222222
2 2
1 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 4
1 1 1
3 342 3
33
1
1
11
1. 1
dt
u du
t t3 4
1 3
1
ttt
III
t t t t
333 333
2 2
1 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 4
1 1 1
3 342 3
33
1
1
11
1. 1
dt
u du
ttt ttt
1 3
1
u u
I du u du u
1
11 12 1 2
2 1 22 23 3
3 3
3
0 0
0
0
1 1 1
13 3 3 2
3
Đặt
dt
u du
333 444
1 3
1
uuu uuu
III ddduuu uuu uuu uuu
3
0 0
0
0
111 111 111
13 3 3 2
3
x
I dx
x x
x
2 2 4
23
1
1
ddd
111
111111 111222 111 222
222 111 222222 222333 333
333 333
3
0 0
0
0
13 3 3 2
3
xxx
III dddxxx
x x
x
222 222
2
333
111
111
Câu 48.
x x
x
444
2
t x2
1 Đặt ttt xxx222
111
http://megabook.vn/
9. t
I dt
t
3 2 2
2
2
( 1)
2
t t
dt t dt dt
t t
3 3 34 2
2
2 2
2 2 2
2 1 1 19 2 4 2
ln
3 4 4 22 2
Trang 9
333 222
ttt
222 222
222 222 222
999
x
I x x dx
x
1
0
1
2 ln 1
1
ttt
t2
2
111
2
ttt ttt ttt ttt
ttt ttt
333 333 333
222 222
222 222 222
111 222 444
lllnnn
333 444 444 222222 222
x
III x x x
1
000
1
222 ln 1
111
x
H dx
x
1
0
1
1
= ddd
222
222111 111
xxx
HHH dx
xxx
111
111
x t tcos ; 0;
2
Dạng 2: Đổi biến số dạng 2
ddd
111
ooo 000;
H 2
2
Câu 49.
xxx
xxx xxx xxx
111
111
lllnnn 111
xxx
dddxxx
000
xxx ttt tttccc sss ;;; ;
222
H 2
2
K x x dx
1
0
2 ln(1 )
Tính
111
ooo 000
222
222
K x x dx
1
0
2 ln(1 )
u x
dv xdx
ln(1 )
2
. Đặt xxx ttt tttccc sss ;;; ;;;
222
HHH
111
000
222 lll )))
(((
K
1
2
KKK xxx xxx dddxxxnnn(((111
uuu xxxlllnnn 111 )))
K
1
2
I x x x dx
2
5 2 2
2
( ) 4
Tính
d
ddd
dd
v
vvvvv
xxx
xxx
(((
222
111
222
x x xdx
2
5 2 2
2
)4
x x x dx
2
5 2 2
2
( ) 4
. Đặt
uuu xxx
dddxxx
lllnnn 111 )))
KKK
I x x x x222 222
222
( ) 4
x
222
222
((( ))) 444
x x dx
2
5 2
2
4
ddd
222
555
xxx x dx222 2
) 4
222
xxx xxx dx555 222
222
444
x x dx
2
2 2
2
4
Câu 50. III xxx xxx xxx xxx((( ))) 444
555
xxx xxx ddd222
dx
2
x x dx2 2
222
4
I = xxx xxx
222
xxx xxx dddxxx555 222
444
222
dx222 222
444
x x dx
2
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 dx
2
2 2
2
4
444 + Tính A = ddd
2
x x d x2 2
2
x t2sin
. Đặt ttt xxx
222
222 222
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 dx
I
x
2 2
4
1
3 4
2
.
III
x x
I
2 2
3 4
x
I dx dx
x x
2 2 2
4 4
1 1
3 4
2 2
Vậy:
ddd
xxx444
111 222
xxx
4 4
111 111
333 444
222 222
.
xxx
222 222 222
I1
Câu 51.
xxx xxx
III
222 222
333 444
III dddxxx dddxxx
444 444
111 dx
x
2
4
1
3
2
Ta có:
xxx
222 222 222
III xxx
222
444 x dx
2
4
1
3 7
2 16
.
ddd
xxx
333
222
2
x dx4
111
3 7
+ Tính xxx
222
111
222
444
222 666
x
I d x
x
2 2
2 4
1
4
2
= ddd
333
111
x
I 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
222 444
xxx dddxxx ttt tttsssiiinnn 222ccc sss . Đặt ttt ddd222 ooo .
http://megabook.vn/
10. tdt
I t dt t d t
t t
22 2 2
2 2
2 4 2
6 6 6
1 cos 1 1 1 3
cot cot . (cot )
8 8 8 8sin sin
I
1
7 2 3
16
Trang 10
tttddd
ddd ttt
ttt
666 666 666
ooo
ooo ccc
888 888 888 888iiinnn iiinnn
666
x dx
I
x
1 2
6
0 4
ttt
III ttt ttt ttt ddd
t t
222222 222 222
222 222
222 444 222
6 6 6
111 ccc sss 111 111 111 333
ccc ttt ooottt ... (((cccooottt )))
8 8 8 8sin sin
III
111666
x x
I
1 2
t x dt x dx3 2
3
Vậy: 111
777 222 333
xxx666
000 444
ttt xxx
dt
I
t
1
2
0
1
3 4
.
ddd
dddxxx333 III
444
Câu 52.
xxx xxx
III
111 222
ttt xxx dddttt xxx dddxxx333 222
dddttt
ttt
111
111
t u u dt udu2sin , 0; 2cos
2
Đặt 333 III
2
22222000
333
t u u dt udu2sin , 0; 2c s
2
I dt
6
0
1
3 18
dddttt111
111
222 ,,, 000 222 sss
222
III ttt
.
ttt uuu uuu dddttt uuuddduuusssiii nnn ;;; ccc ddd
666
111
x
I dx
x
2
0
2
2
Đặt ooo dddttt
666
000
111
333 111888
x t dx tdt2cos 2sin
III
xxx
III
xxx
222
000
222
222
xxx ttt dddxxx ttt ttt222cccsss 222 iiinnn
t
I dt
2
2
0
4 sin 2
2
.
dddxxx dddooo sss
t
I dt
2
2
4 sin 2
Câu 53. xxx ttt dddxxx ttt ttt222cccsss 222 iiinnn
t
t
000
x dx
I
x x
1 2
2
0 3 2
Đặt dddooo sss III dddsss
222
x dx
I
x x2
0 3 2
x dx
I
x
1 2
2 2
0 2 ( 1)
ttt
ttt
222
222
444 iiinnn 222
222
III
xxx222
000 333 222
222 222
000 222 ((( )
x t1 2cos
.
xxx dddxxx
xxx
xxx xxx
III
111 222
xxx ttt111 222ccc sss
Câu 54.
111 222
ddd
xxx 111
ooo
t t
I dt
t
22
2
2
3
(1 2cos ) 2sin
4 (2cos )
Ta có:
xxx xxx
III
111 222
)))
xxx ttt111 222ccc sss
t t
I dddt
t
22
2
2
3
(1 2s ) 2 in
4 (2cos )
t t dt
2
3
2
3 4cos 2cos2
. Đặt ooo
dddttt
ttt
222222
222
222
333
((( 222 ))) 222 iiinnn
444 ((( )))
ttt dddttt
222
333
cccooosss
3 3
4
2 2
.
ttt ttt
III
111 sss
222cccooosss
ttt
222
333 444cccooosss 222 222
3 3
4
x x dx
1
2
2
0
1 2 1
ddd
cccooo sss
ttt dddttt
222
333
cccooosss
333 333
222 222
x dx
0
1 2 1 x tsin
= ttt333 444cccooosss 222 222
444
000
111 222 111 xxx tttsssiiinnn I t t tdt
6
0
3 1
(cos sin )cos
12 8 8
=
xxx dddxxx I t t tdt
6
000
3 1
(cos sin )cos
12 8 8
Câu 55. xxx
111
222
222
xxx tttsssiiinnn ttt )))s
222 888
I x dx
3
2
2
1
Đặt tttdddtttooo nnn cccsss
111
I d
3
2
1
III ttt
666
333 111
(((ccc sss sssiii ooo
888
333
222
111
Dạng 3: Tích phân từng phần
III dddCâu 56. xxx xxx222
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11. x
du dxu x
xdv dx
v x
2
21
1
x
I x x x dx x dx
x x
3 3
2 2
2 2
2 2
3 1
1 . 5 2 1
2 1 1
dx
x dx
x
3 3
2
2
2 2
5 2 1
1
I x x2 3
2
5 2 ln 1
Trang 11
xxx
ddd dddxxxxxx
dv dx
v x
222111
x
I x x x dx x dx
x x
3 3
2 2
2 2
2 2
3 1
1 . 5 2 1
2 1 1
dx
x dx
x
3 3
2
2
2 2
5 2 1
1
I x x2 3
2
5 2 ln 1
I
5 2 1
ln 2 1 ln2
2 4
Đặt
uuuuuu
xxxdv dx
v x
222
111
x
I x x x dx x dx
x x
3 3
2 2
2 2
2 2
3 1
1 . 5 2 1
2 1 1
dx
x dx
x
3 3
2
2
2 2
5 2 1
1
I x x2 3
2
5 2 ln 1
I
5 2 1
ln 2 1 ln2
222
x
t
1
cos
I
5 2 1
ln 2 1 ln2
444
tttcos
2;3 1;1 Chú ý: Không được dùng phép đổi biến xxx
111
cos
;;;333 111;;; vì 222 111
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12. Trang 12
TP3: TÍCH PHÂN HÀM SỐ LƯỢNG GIÁC
x x
I dx
x x
2
8cos sin2 3
sin cos
Dạng 1: Biến đổi lượng giác
xxx xxx
III xxx
x x
222
888cccsss sssiiinnn222 333
sin cos
x x x
I dx x x x x dx
x x
2
(sin cos ) 4cos2
sin cos 4(sin cos
sin cos
x x C3cos 5sin
Câu 57. ddd
x x
ooo
sin cos
xxx xxx
III dddxxx xxx xxx xxx xxx xxx
x x
222
iiinnn cccooosss ))) 444cccooo222
sssiiinnn cccooosss sssiiinnn sss
sin cos
x x C3cos 5sin
xxx
ddd
x x
(((sss sss
444((( cccooo
sin cos
x x C3cos 5sin
x x x
I dx
x
cot tan 2tan2
sin4
.
ddd
x
ooo tttaaa tttaaa
sin4
x x x x
I dx dx dx C
x x xx2
2cot2 2tan2 2cot 4 cos4 1
2
sin4 sin4 2sin4sin 4
Câu 58.
xxx xxx xxx
III xxx
x
ccc ttt nnn 222 nnn222
sin4
xxx
III ddd ddd dddxxx
x x xx2
222cccooo tttaaa cccooo cccooosss
222
sin4 sin4 2sin4sin 4
x
I dx
x x
2
cos
8
sin2 cos2 2
Ta có:
xxx xxx xxx
xxx xxx CCC
x x xx2
ttt222 222 nnn222 222 ttt444 444 111
sin4 sin4 2sin4sin 4
ddd
x x
222
cccooo
sin2 cos2 2
x
I dx
x
1 cos 2
1 4
2 2 1 sin 2
4
x
dx
dx
x x x
2
cos 2
1 4
2 2 1 sin 2 sin cos4 8 8
x
dx
dx
x x2
cos 2
1 14
2 32 2 1 sin 2 sin
4 8
x x C
1 3
ln 1 sin 2 cot
4 84 2
Câu 59.
xxx
III xxx
x x
sss
888
sin2 cos2 2
xxx
III dddxxx
x
111 cccooosss 222
111 444
2 2 1 sin 2
4
x
dx
dx
x x x
2
cos 2
1 4
2 2 1 sin 2 sin cos4 8 8
x
dx
dx
x x2
cos 2
1 14
2 32 2 1 sin 2 sin
4 8
x x C
1 3
ln 1 sin 2 cot
4 84 2
dx
I
x x
3
2 3sin cos
Ta có:
x2 2 1 sin 2
4
x
dx
dx
x x x
2
cos 2
1 4
2 2 1 sin 2 sin cos4 8 8
x
dx
dx
x x2
cos 2
1 14
2 32 2 1 sin 2 sin
4 8
x x C
1 3
ln 1 sin 2 cot
4 84 2
dx
I
xxx
333
dx
I
x
3
1
2
1 cos
3
Câu 60.
dx
I
xxx222 333sssiiinnn cccooosss
x
3
111
2
1 cos
3
dx
I
x2
3
1
4
2sin
2 6
dddxxx
III
x
3
2
1 cos
3
xxx
III
2
3
111
4
2sin
2 6
1
4 3
=
ddd
xxx2
3
4
2sin
2 6
111
4 3
=
4 3
I dx
x
6
0
1
2sin 3
.
ddd
x0 2sin 3
Câu 61. III xxx
x
666
0
111
2sin 3
http://megabook.vn/
13. I dx dx
x x
6 6
0 0
1
1 1 2
2
sin sin sin sin
3 3
x x
dx dx
x x
x
6 6
0 0
coscos 2 6 2 63
sin sin 2cos .sin
3 2 6 2 6
x x
dx dx
x x
6 6
0 0
cos sin
2 6 2 61 1
2 2
sin cos
2 6 2 6
x x
6 6
0 0
ln sin ln cos .....
2 6 2 6
Trang 13
I
I x x x x dx
2
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 cos
cos 2
6 2
6 3
sin 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
Câu 62. I x x x x dx
2
4 4 6 6
(sin cos )(sin cos )
x x x x4 4 6 6
(sin cos )(sin cos ) x x
33 7 3
cos4 cos8
64 16 64
.
xxx xxx xxx xxx(((sssiiinnn cccooosss )))(((sssiiinnn cccooosss )))
333
cccooosss sss
64 16 64
I
33
128
Ta có: 444 444 666 666
xxx xxx
333 777 333
444 cccooo 888
64 16 64
333
111 888
III
333
222
I x x x dx
2
4 4
0
cos2 (sin cos )
.
ddd
0
ccc (((sss ooo
I x x d x x d
2 2
2 2
0 0
1 1 1
cos2 1 sin 2 1 sin 2 (sin2 ) 0
2 2 2
Câu 63. III xxx xxx xxx xxx
222
444 444
0
ooosss222 iiinnn ccc sss )))
xxx xxx
0 0
ccc sss sss (((sss
2 2 2
I x x dx
2
3 2
0
(cos 1)cos .
III xxx xxx dddxxx ddd
222 222
222 222
0 0
111 111 111
ooo 222 111 sssiiinnn 222 111 iiinnn 222 iiinnn222 ))) 000
2 2 2
ddd
222
333 222
0
(((ccc ccc
x d x x d x
2 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
222 222 222
555 222
0 0
ccc sssnnn (((sssiii
15
xdx x dx
2 2
2
0 0
1
cos . (1 cos2 ).
2
=
888
15
xdx x dx
2 2
2 1
cos . (1 cos2 ).
222
4
B = xdx x dx
2 2
2
000 000
1
cos . (1 cos2 ).
4
8
15
=
4
15 4
Vậy I =
888
15 444
–
2
2
0
I cos cos 2x xdx
.
222
222
0
III cccooo cccoooxxx xxxddd
I x xdx x xdx x x dx
2 2 2
2
0 0 0
1 1
cos cos2 (1 cos2 )cos2 (1 2cos2 cos4 )
2 4
Câu 65.
0
sss sss222 xxx
xxx xxxxxx xxxdddxxx xxx xxx ddd
222 222
222
0 0 0
ooosss cccooo222 (((111 cccooo222 sss 111 cccooo222 cccooo 444
2 4
III ddd xxx xxx
222
0 0 0
111 111
ccc sss sss )))cccooo 222 ((( 222 sss sss )))
2 4
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14. x x x
2
0
1 1
( sin2 sin4 )
4 4 8
x
I dx
x
3
2
0
4sin
1 cos
Trang 14
xxx
222
000
((( sssiiinnn iii
888
xxx
dddxxx
x
333
0
444sssiiinnn
1 cos
x x x
x x x x x
x x
3 3
2
4sin 4sin (1 cos )
4sin 4sin cos 4sin 2sin2
1 cos sin
I x x dx2
0
(4sin 2sin2 ) 2
Câu 66. III
x
222
0 1 cos
x x2
sss 444
444sss 444 ccc sss
1 cos sin
I x x dx2
0
(4sin 2sin2 ) 2
I xdx
2
0
1 sin
xxx xxx xxx
xxx xxx xxx xxx xxx
x x
333 333
2
444 iiinnn sssiiinnn (((111 cccooosss )))
iiinnn sssiiinnn ooosss 444sssiiinnn 222 iiinnn222
1 cos sin
I x x dx2
0
(4sin 2sin2 ) 2
III xxxdddxxx
222
0
1 sin
x x x x
I dx dx
22 2
0 0
sin cos sin cos
2 2 2 2
x
dx
2
0
2 sin
2 4
x x
dx dx
3
22
30
2
2 sin sin
2 4 2 4
4 2
Câu 67.
0
1 sin
III xxx ddd
222222 222
0 0
sssiii ccc sss
2 2 2 2
ddd
222
0
sss
x x
dx dx
3
22
30
2
2 sin sin
2 4 2 4
4 2
dx
I
x
4
6
0 cos
xxx xxx xxx xxx
ddd xxx
0 0
nnn ooosss iiinnn cccooosss
2 2 2 2
xxx
xxx
0
222 iiinnn
222 444
x x
dx dx
3
22
30
2
2 sin sin
2 4 2 4
4 2
x6
0 cos
I x x d x
4
2 4
0
28
(1 2tan tan ) (tan )
15
Câu 68.
dddxxx
III
x
444
6
0 cos
444
222 444
0
aaa aaa aaa
111
Ta có: III xxx xxx ddd xxx
0
222888
(((111 222ttt nnn ttt nnn ))) (((ttt nnn )))
555
.
xdx
I
x x
sin2
3 4sin cos2
Dạng 2: Đổi biến số dạng 1
xxxddd
3 4sin cos2
x x
I dx
x x2
2sin cos
2sin 4sin 2
Câu 69.
xxx
III
xxx xxx
sssiiinnn222
3 4sin cos2
xxx xxx
III 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
1
ln sin 1
sin 1
. Đặt sss III xxx CCC
x
111
nnn sssiiinnn 111
sssiiinnn 111
dx
I
x x3 5
sin .cos
x
lll
dddxxx
III
3
x x5
sin .cos
xx
dx
xxx
dx
I 23233
cos.2sin
8
cos.cos.sin
Câu 70.
3
x x5
sin .cos
xxxx 222333222333333
ccc.2sincos.cos.sin
t xtan
xxxx
dddxxx
xxx
dddxxx
III
ooosss.2sin
888
cos.cos.sin
tttaaa I t t t dt x x x C
t x
3 3 4 2
2
3 1 3 1
3 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
333 111 333 111
333 tttaaannn aaannn 333lllnnn ttt nnn
4 2 2tan
t
x
t2
2
sin2
1
. ddd
t x2
ttt aaa
4 2 2tan
ttt
xxx
t2
222
iiinnn222
1
Chú ý:
t2
sss
1
.
http://megabook.vn/
15. dx
I
x x3
sin .cos
Trang 15
dddxxx
III
x x3
sin .cos
dx dx
I
x x x x x2 2
2
sin .cos .cos sin2 .cos
Câu 71.
x x3
sin .cos
xxx xxx
III
x x x x x2 2
222
sin .cos .cos sin2 .cos
t xtan
dx t
dt x
x t2 2
2
; sin2
cos 1
dt t
I dt
t t
t
2
2
1
2
2
1
t x
t dt t C x C
t
2 2
1 tan
( ) ln ln tan
2 2
ddd ddd
x x x x x2 2
sin .cos .cos sin2 .cos
tttaaa
dddxxx ttt
dddttt xxx
2
x t2
222
;;; sssiiinnn222
cos 1
dt t
I dt
t t
t
2
2
1
2
2
1
t x
t dt t C x C
t
2 2
1 tan
( ) ln ln tan
2 2
x x
I xdx
x
2011 2011 2009
5
sin sin
cot
sin
. Đặt ttt xxxnnn
2
x t2
cos 1
dt t
I dt
t t
t
2
2
1
2
2
1
t x
t dt t C x C
t
2 2
1 tan
( ) ln ln tan
2 2
xxx xxx
III dddxxx
x
111 000111 000000
5
iiinnn iiinnn
ccc ttt
sin
xxI xdx xdx
x x
2011 2011 22
4 4
1
1
cotsin cot cot
sin sin
Câu 72. xxx
x
222000 111 222 111 222 999
5
sss sss
ooo
sin
III xxxdddxxx xxxddd
x x
111 222000111111 222222
4 4
111
tttiii ttt
sin sin
t xcot
Ta có:
xxxxxx xxx
x x
222000111
4 4
111
cccooosssnnn cccooo cccooottt
sin sin
ccc I t tdt t t C
2 4024 8046
22011 2011 20112011 2011
t (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 cot
4024 8046
ttt ttt ttt ttt CCC
222 444 666
222222 111 111 111222 111 111
111 )))
444 444 666
xxx xxx CCC
444000222 888000444
000111 222000111000111 222000111
cccooottt cccooo
4024 8046
x x
I dx
x
2
0
sin2 .cos
1 cos
=
444 666
222 111 111222 111 111
ttt
4024 8046
xxx xxx
III xxx
x
222
0
sssiiinnn ...cccooosss
1 cos
x x
I dx
x
22
0
sin .cos
2
1 cos
Câu 73. ddd
x0
222
1 cos
x0
sss
ccc
t x1 cos Ta có:
xxx xxx
III dddxxx
x
222222
0
iiinnn ...cccooosss
222
111 ooosss
ttt xxx111 ooosss
t
I dt
t
2 2
1
( 1)
2 2ln2 1
. Đặt ccc
ttt
III dddttt
t
222 222
1
((( 111)))
222 222lllnnn222 111
I x xdx
3
2
0
sin tan
t1
xxxddd
0
tttaaa
x x x
I x dx dx
x x
23 3
2
0 0
sin (1 cos )sin
sin .
cos cos
Câu 74. III xxx xxx
333
222
0
sssiiinnn nnn
xxx xxx
xxx dddxxx dddxxx
x x
222
0 0
111 ooosss )))sssiiinnn
nnn ...
cccsss ccc
t xcos Ta có:
xxx
III
x x
222333 333
0 0
sssiiinnn ((( ccc
sssiii
ooo ooosss
xxxccc
u
I du
u
1
22
1
1 3
ln2
8
. Đặt ttt ooosss
ddd
u
222
1
I x x dx2
2
sin (2 1 cos2 )
uuu
III uuu
u
111
222
1
111 333
lllnnn222
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/
16. 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
2
2 sin (sin )
3
I
2
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
dx
I
x x
3
2 4
4
sin .cos
+ KKK xxx xxx xxx xxx
2 2
sin 2cos 2 sin cos
2
222
222 iii
333
I
2
2 3
dx
I
xxx xxx
3
sssiiinnn ...ccc sss
dx
I
x x
3
2 2
4
4.
sin 2 .cos
Câu 76.
dx
I
3
222 444
444
ooo
dddxxx
III
2
x x2
4
...
sin 2 .cos
t xtan
x x
333
2 2
4
444
sin 2 .cos
tttaaa
dx
dt
x2
cos
. Đặt ttt xxxnnn dddttt
x2
cos
t dt t
I t dt t
tt t
3
3 32 2 3
2
2 2
11 1
(1 ) 1 1 8 3 4
2 2
3 3
dddxxx
x2
cos
t dt t
I t dt t
tt t
3
3 32 2 3
2
2 2
11 1
(1 ) 1 1 8 3 4
2 2
3 3
2
2
0
sin 2
2 sin
x
I dx
x
.
t dt t
I t dt t
tt t
3
3 32 2 3
2
2 2
11 1
(1 ) 1 1 8 3 4
2 2
3 3
2
2
sin 2
sssiii
x
I dx
x x x
I dx dx
x x
2 2
2 2
0 0
sin2 sin cos
2
(2 sin ) (2 sin )
Câu 77.
2
2
000
sin 2
222 nnn
x
I dx
xxx
ddd ddd
x x2 2
0 0
sss
(2 sin ) (2 sin )
t x2 sin Ta có:
xxx xxx xxx
III xxx xxx
x x
222 222
2 2
0 0
sssiiinnn222 iiinnn cccooosss
222
(2 sin ) (2 sin )
ttt xxx222 sssiiinnn . Đặt
t
I dt dt t
t tt t
33 3
2 2
2 2 2
2 1 2 2
2 2 2 ln
3 2
2ln
2 3
.
ttt
ddd ddd ttt
t tt t
333 333
2 2
2 2 2
222 222
222 222 222 lll
333 222
222
x
I dx
x
6
0
sin
cos2
III ttt ttt
t tt t
333
2 2
2 2 2
111 222
nnn
222lllnnn
333
ddd
x0
cos2
x x
I dx dx
x x
6 6
2
0 0
sin sin
cos2 2cos 1
Câu 78.
xxx
III xxx
x
666
0
sssiiinnn
cos2
xxx xxx
III dddxxx dddxxx
x x2
0 0
sssnnn sssnnn
cos2 2cos 1
t x dt xdxcos sin
x x
666 666
2
0 0
iii iii
cos2 2cos 1
ttt xxx dddttt xxx xxxcccooosss sssiiinnn
x t x t
3
0 1;
6 2
. Đặt ddd
xxx ttt xxx ttt
333
000 111;;;
666 222
t
I dt
tt
3
1
2
2
31
2
1 1 2 2
ln
2 2 2 22 1
Đổi cận:
ttt
ddd
tt
333
111
222
2
31
2
111 111 222
lll
2 2 2 22 1
1 3 2 2
ln
2 2 5 2 6
Ta được III ttt
tt2
31
2
222
nnn
2 2 2 22 1
333 222
2 2 5 2 6
=
111 222
lllnnn
2 2 5 2 6
http://megabook.vn/
17. Trang 17
x
I e x x dx
22
sin 3
0
.sin .cos .
Câu 79. x
I e x x dx
22
sin 3
0
.sin .cos .
t x2
sin Đặt t x2
sin t
e t dt
1
0
1
(1 )
2
e
1
1
2
= e
1
1
2
.
I x x dx
2 12sin sin
2
6
Câu 80. I x x dx
2 12sin sin
2
6
t xcos Đặt t xcos I
3
( 2)
16
. I
3
( 2)
16
x
I dx
x x
4
6 6
0
sin4
sin cos
Câu 81.
x
I dx
x x
4
6 6
0
sin4
sin cos
x
I dx
x
4
20
sin4
3
1 sin 2
4
x
I dx
x
4
20
sin4
3
1 sin 2
4
t x23
1 sin 2
4
. Đặt t x23
1 sin 2
4
dt
t
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
.
x
I dx
x x
2
3
0
sin
sin 3cos
Câu 82.
x
I dx
x x
2
3
0
sin
sin 3cos
x x xsin 3cos 2cos
6
Ta có: x x xsin 3cos 2cos
6
x xsin sin
6 6
;
x xsin sin
6 6
x x
3 1
sin cos
2 6 2 6
= x x
3 1
sin cos
2 6 2 6
x dx
dx
x x
2 2
3 20 0
sin
63 1
16 16
cos cos
6 6
I =
x dx
dx
x x
2 2
3 20 0
sin
63 1
16 16
cos cos
6 6
3
6
=
3
6
x x
I dx
x
24
2
3
sin 1 cos
cos
Câu 83.
x x
I dx
x
24
2
3
sin 1 cos
cos
x x
I x dx x dx
x x
4 4
2
2 2
3 3
sin sin
1 cos . sin
cos cos
x x
x dx x dx
x x
0 4
2 2
0
3
sin sin
sin sin
cos cos
x x
I x dx x dx
x x
4 4
2
2 2
3 3
sin sin
1 cos . sin
cos cos
x x
x dx x dx
x x
0 4
2 2
0
3
sin sin
sin sin
cos cos
x x
dx dx
x x
0 2 24
2 2
0
3
sin sin
cos cos
7
3 1
12
=
x x
dx dx
x x
0 2 24
2 2
0
3
sin sin
cos cos
7
3 1
12
.
http://megabook.vn/
18. Trang 18
I dx
x x
6
0
1
sin 3cos
Câu 84. I dx
x x
6
0
1
sin 3cos
I dx
x x
6
0
1
sin 3cos
I dx
x x
6
0
1
sin 3cos
dx
x
6
0
1 1
2
sin
3
= dx
x
6
0
1 1
2
sin
3
x
dx
x
6
20
sin
1 3
2
1 cos
3
=
x
dx
x
6
20
sin
1 3
2
1 cos
3
.
t x dt x dxcos sin
3 3
Đặt t x dt x dxcos sin
3 3
I dt
t
1
2
2
0
1 1 1
ln3
2 41
I dt
t
1
2
2
0
1 1 1
ln3
2 41
I x xdx
2
2
0
1 3sin2 2cos
Câu 85. I x xdx
2
2
0
1 3sin2 2cos
I x x dx
2
0
sin 3cos
I x x dx
2
0
sin 3cos
I x x dx x x dx
3 2
0
3
sin 3cos sin 3cos
3 3 = I x x dx x x dx
3 2
0
3
sin 3cos sin 3cos
3 3
xdx
I
x x
2
3
0
sin
(sin cos )
Câu 86.
xdx
I
x x
2
3
0
sin
(sin cos )
x t dx dt
2
Đặt x t dx dt
2
tdt xdx
I
t t x x
2 2
3 3
0 0
cos cos
(sin cos ) (sin cos )
tdt xdx
I
t t x x
2 2
3 3
0 0
cos cos
(sin cos ) (sin cos )
dx dx
2I x
x x x
2 2 4
2
2 00 0
1 1
cot( ) 1
2 2 4(sin cos ) sin ( )
4
dx dx
2I x
x x x
2 2 4
2
2 00 0
1 1
cot( ) 1
2 2 4(sin cos ) sin ( )
4
I
1
2
I
1
2
x x
I dx
x x
2
3
0
7sin 5cos
(sin cos )
Câu 87.
x x
I dx
x x
2
3
0
7sin 5cos
(sin cos )
xdx xdx
I I
x x x x
2 2
1 23 3
0 0
sin cos
;
sin cos sin cos
Xét:
xdx xdx
I I
x x x x
2 2
1 23 3
0 0
sin cos
;
sin cos sin cos
.
x t
2
Đặt x t
2
. Ta chứng minh được I1 = I2
dx dx
x
x x x
2 2
2
20 0
1
tan( ) 122 4sin cos 02cos ( )
4
Tính I1 + I2 =
dx dx
x
x x x
2 2
2
20 0
1
tan( ) 122 4sin cos 02cos ( )
4
http://megabook.vn/
19. 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 x
I dx
x x
2
3
0
3sin 2cos
(sin cos )
Câu 88.
x x
I dx
x x
2
3
0
3sin 2cos
(sin cos )
x t dx dt
2
Đặt x t dx dt
2
t t x x
I dt dx
t t x x
2 2
3 3
0 0
3cos 2sin 3cos 2sin
(cos sin ) (cos sin )
t t x x
I dt dx
t t x x
2 2
3 3
0 0
3cos 2sin 3cos 2sin
(cos sin ) (cos sin )
x x x x
I I I dx dx dx
x x x x x x
2 2 2
3 3 2
0 0 0
3sin 2cos 3cos 2sin 1
2 1
(sin cos ) (cos sin ) (sin cos )
x x x x
I I I dx dx dx
x x x x x x
2 2 2
3 3 2
0 0 0
3sin 2cos 3cos 2sin 1
2 1
(sin cos ) (cos sin ) (sin cos )
I
1
2
I
1
2
.
x x
I dx
x2
0
sin
1 cos
Câu 89.
x x
I dx
x2
0
sin
1 cos
t t t
x t dx dt I dt dt I
t t2 2
0 0
( )sin sin
1 cos 1 cos
t d t
I dt I
t t
2
2 2
0 0
sin (cos )
2
4 4 81 cos 1 cos
Đặt
t t t
x t dx dt I dt dt I
2
t t2
0 0
( )sin sin
1 cos 1 cos
t d t
I dt I
t t
2
2 2
0 0
sin (cos )
2
4 4 81 cos 1 cos
x x
I dx
x x
42
3 3
0
cos sin
cos sin
Câu 90.
x x
I dx
x x
42
3 3
0
cos sin
cos sin
x t dx dt
2
Đặt x t dx dt
2
t t x x
I dt dx
t t x x
0 4 42
3 3 3 3
0
2
sin cos sin cos
cos sin cos sin
t t x x
I dt dx
t t x x
0 4 42
3 3 3 3
0
2
sin cos sin cos
cos sin cos sin
x x x x x x x x
I dx dx xdx
x x x x
4 4 3 32 2 2
3 3 3 3
0 0 0
cos sin sin cos sin cos (sin cos ) 1 1
2 sin2
2 2sin cos sin cos
x x x x x x x x
I dx dx xdx
x x x x
4 4 3 32 2 2
3 3 3 3
0 0 0
cos sin sin cos sin cos (sin cos ) 1 1
2 sin2
2 2sin cos sin cos
I
1
4
I
1
4
.
I x dx
x
2
2
2
0
1
tan (cos )
cos (sin )
Câu 91. I x dx
x
2
2
2
0
1
tan (cos )
cos (sin )
x t dx dt
2
Đặt x t dx dt
2
I t dt
t
2
2
2
0
1
tan (sin )
cos (cos )
x dx
x
2
2
2
0
1
tan (sin )
cos (cos )
I t dt
t
2
2
2
0
1
tan (sin )
cos (cos )
x dx
x
2
2
2
0
1
tan (sin )
cos (cos )
http://megabook.vn/
20. Trang 20
I x x dx
x x
2
2 2
2 2
0
1 1
2 tan (cos ) tan (sin )
cos (sin ) cos (cos )
Do đó: I x x dx
x x
2
2 2
2 2
0
1 1
2 tan (cos ) tan (sin )
cos (sin ) cos (cos )
dt
2
0
2
= dt
2
0
2
I
2
I
2
.
x x
I dx
x
4
0
cos sin
3 sin2
Câu 92.
x x
I dx
x
4
0
cos sin
3 sin2
u x xsin cos
du
I
u
2
2
1 4
Đặt u x xsin cos
du
I
u
2
2
1 4
u t2sin
tdt
I dt
t
4 4
2
6 6
2cos
124 4sin
. Đặt u t2sin
tdt
I dt
t
4 4
2
6 6
2cos
124 4sin
.
x
I dx
x x
3
2
0
sin
cos 3 sin
Câu 93.
x
I dx
x x
3
2
0
sin
cos 3 sin
t x2
3 sin Đặt t x2
3 sin x2
4 cos= x2
4 cos x t2 2
cos 4 . Ta có: 2
x t2
cos 4
x x
dt dx
x2
sin cos
3 sin
và
x x
dt dx
x2
sin cos
3 sin
.
x
dx
x x
3
2
0
sin
.
cos 3 sin
I =
x
dx
x x
3
2
0
sin
.
cos 3 sin
x x
dx
x x
3
2 2
0
sin .cos
cos 3 sin
=
x x
dx
x x
3
2 2
0
sin .cos
cos 3 sin
dt
t
15
2
2
3 4
=
dt
t
15
2
2
3 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 2
ln
4 2
=
t
t
15
2
3
1 2
ln
4 2
1 15 4 3 2
ln ln
4 15 4 3 2
=
1 15 4 3 2
ln ln
4 15 4 3 2
1
ln 15 4 ln 3 2
2
= 1
ln 15 4 ln 3 2
2
.
x x x x
I dx
x x
2
3
3 2
3
( sin )sin
sin sin
Câu 94.
x x x x
I dx
x x
2
3
3 2
3
( sin )sin
sin sin
x dx
I dx
xx
2 2
3 3
2
3 3
1 sinsin
x dx
I dx
xx
2 2
3 3
2
3 3
1 sinsin
.
x
I dx
x
2
3
1 2
3
sin
+ Tính
x
I dx
x
2
3
1 2
3
sin
u x
du dx
dx
dv v x
x2
cot
sin
. Đặt
u x
du dx
dx
dv v x
x2
cot
sin
I1
3
I1
3
dx dx dx
I =
x x
x
2 2 2
3 3 3
2
2
3 3 3
4 2 3
1 sin
1 cos 2cos
2 4 2
+ Tính
dx dx dx
I =
x x
x
2 2 2
3 3 3
2
2
3 3 3
4 2 3
1 sin
1 cos 2cos
2 4 2
I 4 2 3
3
Vậy: I 4 2 3
3
.
http://megabook.vn/
21. Trang 21
x
dx
x x
I
2
2 2
0
sin2
cos 4sin
Câu 95.
x
dx
x x
I
2
2 2
0
sin2
cos 4sin
x x
dx
x
I
2
2
0
2sin cos
3sin 1
x x
dx
x
I
2
2
0
2sin cos
3sin 1
u x2
3sin 1 . Đặt u x2
3sin 1
udu
du
u
I
2 2
1 1
2
2 23
3 3
udu
du
u
I
2 2
1 1
2
2 23
3 3
x
I dx
x
6
0
tan
4
cos2
Câu 96.
x
I dx
x
6
0
tan
4
cos2
x
x
I dx dx
x x
26 6
2
0 0
tan
tan 14
cos2 (tan 1)
x
x
I dx dx
x x
26 6
2
0 0
tan
tan 14
cos2 (tan 1)
t x dt dx x dx
x
2
2
1
tan (tan 1)
cos
. Đặt t x dt dx x dx
x
2
2
1
tan (tan 1)
cos
dt
I
tt
1
1
3
3
2
00
1 1 3
1 2( 1)
dt
I
tt
1
1
3
3
2
00
1 1 3
1 2( 1)
.
x
I dx
x x
3
6
cot
sin .sin
4
Câu 97.
x
I dx
x x
3
6
cot
sin .sin
4
x
I dx
x x
3
2
6
cot
2
sin (1 cot )
x
I dx
x x
3
2
6
cot
2
sin (1 cot )
x t1 cot dx dt
x2
1
sin
. Đặt x t1 cot dx dt
x2
1
sin
t
I dt t t
t
3 1 3 1
3 1
3 1 3
3
1 2
2 2 ln 2 ln 3
3
t
I dt t t
t
3 1 3 1
3 1
3 1 3
3
1 2
2 2 ln 2 ln 3
3
dx
I
x x
3
2 4
4
sin .cos
Câu 98.
dx
I
x x
3
2 4
4
sin .cos
dx
I
x x
3
2 2
4
4.
sin 2 .cos
Ta có:
dx
I
x x
3
2 2
4
4.
sin 2 .cos
dt
t x dx
t2
tan
1
. Đặt
dt
t x dx
t2
tan
1
t dt t
I t dt t
tt t
3
2 2 33 3(1 ) 1 1 8 3 42( 2 ) ( 2 )
2 2 3 31 1 1
t dt t
I t dt t
tt t
3
2 2 33 3(1 ) 1 1 8 3 42( 2 ) ( 2 )
2 2 3 31 1 1
http://megabook.vn/
22. Trang 22
x
I dx
x x x
4
2
0
sin
5sin .cos 2cos
Câu 99.
x
I dx
x x x
4
2
0
sin
5sin .cos 2cos
x
I dx
x x x
4
2 2
0
tan 1
.
5tan 2(1 tan ) cos
Ta có:
x
I dx
x x x
4
2 2
0
tan 1
.
5tan 2(1 tan ) cos
t xtan. Đặt t xtan ,
t
I dt dt
t tt t
1 1
2
0 0
1 2 1 1 2
ln3 ln2
3 2 2 1 2 32 5 2
t
I dt dt
t tt t
1 1
2
0 0
1 2 1 1 2
ln3 ln2
3 2 2 1 2 32 5 2
xdx
x x x
I
24
4 2
4
sin
cos (tan 2tan 5)
Câu 100.
xdx
x x x
I
24
4 2
4
sin
cos (tan 2tan 5)
dt
t x dx
t2
tan
1
Đặt
dt
t x dx
t2
tan
1
t dt dt
I
t t t t
21 1
2 2
1 1
2
2 ln 3
32 5 2 5
t dt dt
I
t t t t
21 1
2 2
1 1
2
2 ln 3
32 5 2 5
dt
I
t t
1
1 2
1 2 5
Tính
dt
I
t t
1
1 2
1 2 5
t
u I du
0
1
4
1 1
tan
2 2 8
. Đặt
t
u I du
0
1
4
1 1
tan
2 2 8
I
2 3
2 ln
3 8
. Vậy I
2 3
2 ln
3 8
.
x
I dx
x
22
6
sin
sin3
Câu 101.
x
I dx
x
22
6
sin
sin3
.
x x
I dx dx
x x x
22 2
3 2
6 6
sin sin
3sin 4sin 4cos 1
x x
I dx dx
x 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
I
t t
3
0 2
2
203
2
1 1
ln(2 3)
14 44 1
4
dt dt
I
t t
3
0 2
2
203
2
1 1
ln(2 3)
14 44 1
4
x x
I dx
x
2
4
sin cos
1 sin2
Câu 102.
x x
I 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 x
I dx
x x
2
4
sin cos
sin cos
x x
I dx
x x
2
4
sin cos
sin cos
t x x dt x x dxsin cos (cos sin )
I dt t
t
22
11
1 1
ln ln2
2
. Đặt t x x dt x x dxsin cos (cos sin )
I dt t
t
22
11
1 1
ln ln2
2
http://megabook.vn/
23. Trang 23
I x x xdx
2
6 3 5
1
2 1 cos .sin .cos Câu 103. I x x xdx
2
6 3 5
1
2 1 cos .sin .cos
t dt
t x t x t dt x xdx dx
x x
5
6 3 6 3 5 2
2
2
1 cos 1 cos 6 3cos sin
cos sin
t t
I t t dt
1
1 7 13
6 6
00
12
2 (1 ) 2
7 13 91
Đặt
t dt
t x t x t dt x xdx dx
x x
5
6 3 6 3 5 2
2
2
1 cos 1 cos 6 3cos sin
cos sin
t t
I t t dt
1
1 7 13
6 6
00
12
2 (1 ) 2
7 13 91
xdx
I
x x
4
2
0
tan
cos 1 cos
Câu 104.
xdx
I
x x
4
2
0
tan
cos 1 cos
xdx
I
x x
4
2 2
0
tan
cos tan 2
Ta có:
xdx
I
x x
4
2 2
0
tan
cos tan 2
2 2 2
2
tan
2 tan 2 tan
cos
x
t x t x tdt dx
x
. Đặt 2 2 2
2
tan
2 tan 2 tan
cos
x
t x t x tdt dx
x
3 3
2 2
3 2
tdt
I dt
t
3 3
2 2
3 2
tdt
I dt
t
x
I dx
x x
2
3
0
cos2
(cos sin 3)
Câu 105.
x
I dx
x x
2
3
0
cos2
(cos sin 3)
t x xcos sin 3 Đặt t x xcos sin 3
t
I dt
t
4
3
2
3 1
32
t
I dt
t
4
3
2
3 1
32
.
x
I dx
x x
4
2 4
0
sin4
cos . tan 1
Câu 106.
x
I dx
x x
4
2 4
0
sin4
cos . tan 1
x
I dx
x x
4
4 4
0
sin4
sin cos
Ta có:
x
I dx
x x
4
4 4
0
sin4
sin cos
t x x4 4
sin cos . Đặt t x x4 4
sin cos I dt
2
2
1
2 2 2 .
x
I dx
x
4
2
0
sin4
1 cos
Câu 107.
x
I dx
x
4
2
0
sin4
1 cos
x x
I dx
x
24
2
0
2sin2 (2cos 1)
1 cos
Ta có:
x x
I dx
x
24
2
0
2sin2 (2cos 1)
1 cos
t x2
cos. Đặt t x2
cos
t
I dt
t
1
2
1
2(2 1) 1
2 6ln
1 3
t
I dt
t
1
2
1
2(2 1) 1
2 6ln
1 3
.
x
I dx
x
6
0
tan( )
4
cos2
Câu 108.
x
I dx
x
6
0
tan( )
4
cos2
26
2
0
tan 1
(tan 1)
x
I dx
x
Ta có:
26
2
0
tan 1
(tan 1)
x
I dx
x
t xtan. Đặt t xtan
1
3
2
0
1 3
( 1) 2
dt
I
t
1
3
2
0
1 3
( 1) 2
dt
I
t
.
http://megabook.vn/
24. Trang 24
36
0
tan
cos2
x
I dx
x
Câu 109.
36
0
tan
cos2
x
I dx
x
3 36 6tan tan
2 2 2 2cos sin cos (1 tan )0 0
x x
I dx dx
x x x x
Ta có:
3 36 6tan tan
2 2 2 2cos sin cos (1 tan )0 0
x x
I dx dx
x x x x
.
t xtanĐặt t xtan
3
33 1 1 2
ln
2 6 2 310
t
I dt
t
3
33 1 1 2
ln
2 6 2 310
t
I dt
t
.
x
I dx
x
2
0
cos
7 cos2
Câu 110.
x
I dx
x
2
0
cos
7 cos2
xdx
I
x
2
2 2
0
1 cos
2 6 22 sin
xdx
I
x
2
2 2
0
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
x
x
x
3
3
84
4 3
1
sin
.cos
cos
dx
xx
3
24 3
4
1 1
.
costan
Ta có: dx
x
x
x
3
3
84
4 3
1
sin
.cos
cos
dx
xx
3
24 3
4
1 1
.
costan
.
t xtanĐặt t xtan I t dt
33
84
1
4 3 1
I t dt
33
84
1
4 3 1
3
2
0
cos cos sin
( )
1 cos
x x x
I x dx
x
Câu 112.
3
2
0
cos cos sin
( )
1 cos
x x x
I x dx
x
x x x x x
I x dx x x dx dx J K
x x
2
2 2
0 0 0
cos (1 cos ) sin .sin
.cos .
1 cos 1 cos
Ta có:
x x x x x
I x dx x x dx dx J K
x x
2
2 2
0 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 x
K dx
x2
0
.sin
1 cos
+ Tính
x x
K dx
x2
0
.sin
1 cos
x t dx dt
t t t t x x
K dt dt dx
t t x2 2 2
0 0 0
( ).sin( ) ( ).sin ( ).sin
1 cos ( ) 1 cos 1 cos
x x x x dx x dx
K dx K
x x x2 2 2
0 0 0
( ).sin sin . sin .
2
21 cos 1 cos 1 cos
. Đặt x t dx dt
t t t t x x
K dt dt dx
t t x2 2 2
0 0 0
( ).sin( ) ( ).sin ( ).sin
1 cos ( ) 1 cos 1 cos
x x x x dx x dx
K dx K
2
x x x2 2
0 0 0
( ).sin sin . sin .
2
21 cos 1 cos 1 cos
http://megabook.vn/
25. Trang 25
t xcos
dt
K
t
1
2
1
2 1
Đặt t xcos
dt
K
t
1
2
1
2 1
t u dt u du2
tan (1 tan )
u du
K du u
u
2 24 4
4
2
4
4 4
(1 tan )
.
2 2 2 41 tan
, đặt t u dt u du2
tan (1 tan )
u du
K du u
u
2 24 4
4
2
4
4 4
(1 tan )
.
2 2 2 41 tan
I
2
2
4
Vậy I
2
2
4
2
2
6
cos
I
sin 3 cos
x
dx
x x
Câu 113.
2
2
6
cos
I
sin 3 cos
x
dx
x x
2
2 2
6
sin cos
sin 3 cos
x x
I dx
x x
Ta có:
2
2 2
6
sin cos
sin 3 cos
x x
I dx
x x
t x2
3 cos . Đặt t x2
3 cos
dt
I
t
15
2
2
3
1
ln( 15 4) ln( 3 2)
24
dt
I
t
15
2
2
3
1
ln( 15 4) ln( 3 2)
24
Dạng 3: Đổi biến số dạng 2
I x x dx
2 12sin sin .
2
6
Câu 114. I x x dx
2 12sin sin .
2
6
x t t
3
cos sin , 0
2 2
Đặt x t t
3
cos sin , 0
2 2
tdt
4
2
0
3
cos
2
I = tdt
4
2
0
3
cos
2
3 1
2 4 2
=
3 1
2 4 2
.
2
2 2
0
3sin 4cos
3sin 4cos
x x
I dx
x x
Câu 115.
2
2 2
0
3sin 4cos
3sin 4cos
x x
I 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 dx
x x x
2 2
2 2
0 0
3sin 4cos
3 cos 4 sin
x x
dx 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 dx
x x x
2 2
2 2
0 0
3sin 4cos
3 cos 4 sin
x x
dx dx
x x
2
1 2
0
3sin
3 cos
x
I dx
x
+ Tính
2
1 2
0
3sin
3 cos
x
I dx
x
cos sin t x dt xdx. Đặt cos sin t x dt xdx
1
1 2
0
3
3
dt
I
t
1
1 2
0
3
3
dt
I
t
http://megabook.vn/
26. Trang 26
2
3 tan 3(1 tan ) t u dt u duĐặt 2
3 tan 3(1 tan ) t u dt u du
26
1 2
0
3 3(1 tan ) 3
3(1 tan ) 6
u du
I
u
26
1 2
0
3 3(1 tan ) 3
3(1 tan ) 6
u du
I
u
2
2 2
0
4cos
4 sin
x
I dx
x
+ Tính
2
2 2
0
4cos
4 sin
x
I dx
x
1 1sin cos t x dt xdx
1
1
2 12
10
4
ln3
4
dt
I dt
t
. Đặt 1 1sin cos t x dt xdx
1
1
2 12
10
4
ln3
4
dt
I dt
t
3
ln3
6
IVậy:
3
ln3
6
I
x
I dx
x x
4
2
6
tan
cos 1 cos
Câu 116.
x
I dx
x x
4
2
6
tan
cos 1 cos
x x
I dx dx
x xx
x
4 4
2 2
2
26 6
tan tan
1 cos tan 2cos 1
cos
Ta có:
x x
I dx dx
x xx
x
4 4
2 2
2
26 6
tan tan
1 cos tan 2cos 1
cos
u x du dx
x2
1
tan
cos
Đặt u x du dx
x2
1
tan
cos
u
I dx
u
1
2
1
3
2
u
I dx
u
1
2
1
3
2
u
t u dt du
u
2
2
2
2
. Đặt
u
t u dt du
u
2
2
2
2
I dt t
3
3
7
7 3
3
7 3 7
3 .
3 3
.
I dt t
3
3
7
7 3
3
7 3 7
3 .
3 3
x
I dx
x x
2
4
sin
4
2sin cos 3
Câu 117.
x
I dx
x x
2
4
sin
4
2sin cos 3
x x
I dx
x x
2
2
4
1 sin cos
2 sin cos 2
Ta có:
x x
I 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
2
0
1 1
2 2
I dt
t
1
2
0
1 1
2 2
t u2tanĐặt t u2tan
u
I du
u
1
arctan
22
2
0
1 2(1 tan ) 1 1
arctan
22 22tan 2
u
I du
u
1
arctan
22
2
0
1 2(1 tan ) 1 1
arctan
22 22tan 2
http://megabook.vn/
27. x x
I dx
x
3
2
3
sin
cos
Trang 27
I dx
x
3
2
3
cos
Dạng 4: Tích phân từng phần
III dddxxx
xxx
333
222
333
cccooosss
x dx
I xd J
x x x
3 33
3
3 3
1 4
,
cos cos cos 3
Câu 118.
xxx xxxsssiiinnn
dddxxx
xxx
cccooo cccooo
dx
J
x
3
3
cos
.
xxx
III ddd JJJ
xxx xxx xxx
333 333333
333
333 333
111 444
,,,
sss cccooo sss 333
xxx
t xsin .
Sử dụng công thức tích phân từng phần ta có:
sss
dddxxx
JJJ
333
333
cccooosss
iiinnn ...
d x d t t
J
x tt
3 3
3 2 2
2 3
3
23 2
1 1 2 3
ln ln
cos 2 1 2 31
với
dx
J
x
3
ttt xxxsss
dx dt t
J
ttt
3 3
3 2 2
222
333
222
222
1 1 2 3
ln ln
333111
I
4 2 3
ln .
3 2 3
Để tính J ta đặt
ddd
x
xx
t
tt
333 222 222
333
333
111 222
lllnnn lll
ccc sss 222 111 222
I
4 222 333
lnnn ...
Khi đó
xxx dddttt ttt
JJJ
333 333
111 333
nnn
ooo
333 222 333
xx
I e dx
x
2
0
1 sin
.
1 cos
x x
x x
x xx 2 2
1 2sin cos1 sin 12 2 tan
1 cos 2
2cos 2cos
2 2
Vậy III
444
lll
xxxxxx
III eee dddxxx
222
000
111 sssiiinnn
...
cccooo
x x
xxx x
xxx 2 2
1 sin coss nnn 111222 2 tan
111 222
222 222 sss
222 222
x
xe dx x
I e dx
x
2 2
20 0
tan
2
2cos
2
Câu 119. xx
I e dx
xxx
2
1 sin
.
111 sss
xxx222 222
222111 iii
ccc sss
cccooosss cccooo
eeexxx
xxxxxx xxx
III eee dddxxx
xxx
222 222
tttaaannn
222
e2
Ta có:
xxx xxx
xxx
xxx
111 sssiiinnn cccooossssss 222 tttaaannn
ooo
ddd
222000 000
222 ooosss
222
e2
x x
I dx
x
4
2
0
cos2
1 sin2
ex
xdx x
I e dx
2 2
tan
2
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/
28. Trang 28
u x du dx
x
dv dx v
xx 2
cos2 1
1 sin2(1 sin2 )
Đặt
u x du dx
x
dv dx v
xx 2
cos2 1
1 sin2(1 sin2 )
I x dx dx
x x
x
4 4
20 0
1 1 1 1 1 1 1
. . .4
2 1 sin2 2 1 sin2 16 2 20 cos
4
x
1 1 1 2 2
. tan . 0 14
16 2 4 16 2 2 4 162 0
I x dx dx
x x
x
4 4
20 0
1 1 1 1 1 1 1
. . .4
2 1 sin2 2 1 sin2 16 2 20 cos
4
x
1 1 1 2 2
. tan . 0 14
16 2 4 16 2 2 4 162 0
TP4: TÍCH PHÂN HÀM SỐ MŨ - LOGARIT
Dạng 1: Đổi biến số
x
x
e
I dx
e
2
1
Câu 121.
x
x
e
I dx
e
2
1
x x x
t e e t e dx tdt2
2 Đặt x x x
t e e t e dx tdt2
2
t
I dt
t
3
2
1
t t t t C3 22
2 2ln 1
3
x x x x x
e e e e e C
2
2 2ln 1
3
.
t
I dt
t
3
2
1
t t t t C3 22
2 2ln 1
3
x x x x x
e e e e e C
2
2 2ln 1
3
x
x
x x e
I dx
x e
2
( )
Câu 122.
x
x
x x e
I dx
x e
2
( )
x
x
x x e
I dx
x e
2
( )
x
x
x x e
I dx
x e
2
( )
x x
x
xe x e
dx
xe
.( 1)
1
=
x x
x
xe x e
dx
xe
.( 1)
1
x
t x e. 1 . Đặt x
t x e. 1 x x
I xe xe C1 ln 1 x x
I xe xe C1 ln 1 .
x
dx
I
e2
9
Câu 123.
x
dx
I
e2
9
x
t e2
9 Đặt x
t e2
9
dt t
I C
tt2
1 3
ln
6 39
x
x
e
C
e
2
2
1 9 3
ln
6 9 3
dt t
I C
tt2
1 3
ln
6 39
x
x
e
C
e
2
2
1 9 3
ln
6 9 3
x
x
x x
I dx
ex e
2
2
2 1
ln(1 ) 2011
ln ( )
Câu 124.
x
x
x x
I dx
ex e
2
2
2 1
ln(1 ) 2011
ln ( )
x x
I dx
x x
2
2 2
ln( 1) 2011
( 1) ln( 1) 1
Ta có:
x x
I dx
x x
2
2 2
ln( 1) 2011
( 1) ln( 1) 1
t x2
ln( 1) 1 . Đặt t x2
ln( 1) 1
t
I dt
t
1 2010
2
t t C
1
1005ln
2
t
I dt
t
1 2010
2
t t C
1
1005ln
2
x x C2 21 1
ln( 1) 1005ln(ln( 1) 1)
2 2
= x x C2 21 1
ln( 1) 1005ln(ln( 1) 1)
2 2
http://megabook.vn/
29. Trang 29
e x
x
xe
J dx
x e x1
1
( ln )
Câu 125.
e x
x
xe
J dx
x e x1
1
( ln )
e x ee
x
x
d e x e
J e x
ee x 11
( ln ) 1
ln ln ln
ln
e x ee
x
x
d e x e
J e x
ee x 11
( ln ) 1
ln ln ln
ln
x x
x x x
e e
I dx
e e e
ln2 3 2
3 2
0
2 1
1
Câu 126.
x x
x x x
e e
I dx
e e e
ln2 3 2
3 2
0
2 1
1
x x x x x x
x x x
e e e e e e
I dx
e e e
ln2 3 2 3 2
3 2
0
3 2 ( 1)
1
x x x x x x
x x x
e e e e e e
I dx
e e e
ln2 3 2 3 2
3 2
0
3 2 ( 1)
1
x x x
x x x
e e e
dx
e e e
ln2 3 2
3 2
0
3 2
1
1
=
x x x
x x x
e e e
dx
e e e
ln2 3 2
3 2
0
3 2
1
1
x x x
e e e x3 2 ln2 ln2
ln( – 1)
0 0
= x x x
e e e x3 2 ln2 ln2
ln( – 1)
0 0
14
ln
4
= ln11 – ln4 =
14
ln
4
x
dx
I
e
3ln2
2
30
2
Câu 127.
x
dx
I
e
3ln2
2
30
2
x
x
x
e dx
I
e e
3ln2 3
2
0 33 2
x
x
x
e dx
I
e e
3ln2 3
2
0 33 2
x x
t e dt e dx3 31
3
. Đặt
x x
t e dt e dx3 31
3
I
3 3 1
ln
4 2 6
I
3 3 1
ln
4 2 6
x
I e dx
ln2
3
0
1 Câu 128. x
I e dx
ln2
3
0
1
x
e t
3
1
t dt
dx
t
2
3
3
1
t dt
dx
t
2
3
3
1
dt
t
1
3
0
1
3 1
1
I = dt
t
1
3
0
1
3 1
1
dt
t
1
3
0
3 3
1
=
dt
t
1
3
0
3 3
1
.
dt
I
t
1
1 3
0
3
1
Tính
dt
I
t
1
1 3
0
3
1
t
dt
t t t
1
2
0
1 2
1 1
=
t
dt
t t t
1
2
0
1 2
1 1
ln2
3
= ln2
3
I 3 ln2
3
Vậy: I 3 ln2
3
x x
x x x x
e e dx
I
e e e e
ln15 2
3ln2
24
1 5 3 1 15
Câu 129.
x x
x x x x
e e dx
I
e e e e
ln15 2
3ln2
24
1 5 3 1 15
x x
t e t e2
1 1 x
e dx tdt2 Đặt x x
t e t e2
1 1 x
e dx tdt2
t t dt
I dt t t t
t tt
4 42 4
2 3
3 3
(2 10 ) 3 7
2 2 3ln 2 7ln 2
2 24
2 3ln2 7ln6 7ln5
.
t t dt
I dt t t t
t tt
4 42 4
2 3
3 3
(2 10 ) 3 7
2 2 3ln 2 7ln 2
2 24
2 3ln2 7ln6 7ln5
ln3 2
ln2 1 2
x
x x
e dx
I
e e
Câu 130.
ln3 2
ln2 1 2
x
x x
e dx
I
e e
x
e 2 x
e dx tdt2
2 Đặt t = x
e 2 x
e dx tdt2
2
t tdt
t t
1 2
2
0
( 2)
1
I = 2
t tdt
t t
1 2
2
0
( 2)
1
t
t dt
t t
1
2
0
2 1
1
1
= 2
t
t dt
t t
1
2
0
2 1
1
1
t dt
1
0
2 ( 1)=
1
t dt
0
2 ( 1)
d t t
t t
1 2
2
0
( 1)
2
1
+
d t t
t t
1 2
2
0
( 1)
2
1
t t
1
2
0( 2 )= t t
1
2
0( 2 ) t t
1
2
02ln( 1) + t t
1
2
02ln( 1) 2ln3 1= 2ln3 1 .
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