Thomas Calculus 13th [Solutions]

648 Chapter 8 Techniques of Integration
102.
2
1 x
(1 x)
3
dx;
u
du
1 x
dx
2 2
1 ( u 1) u 2 u 2 1 2 2 ln | | 2 1
3 3 u
2 3 u 2
u u u u u
du du du du du u C
ln |1 x |
2 1
1 x (1 x)
2
C
103. x 1 x dx;
2
w x w x 2
2w
2w dw dx
1 w dw
1 w
2 ( ) 2 3/2
2 w
(1 w)
3
( ) 4 5/2
4 w
(1 w)
15
4
( ) 8 7/2
(1 w)
105
2 2 3/2 5/2 7/2
0
2w 1 w dw
4
w (1 w) 16
w(1 w) 32 (1 w)
3 15 105
C
3/2
16
5/2
32
7/2
4
x 1 x x 1 x 1 x
3 15 105
C
104. 1 1 x dx;
1
2
1
2
w x w x
w dw dx
2w
1 w dw;
2 3/2
u 2 w, du 2 dw, dv 1 w dw, v (1 w)
3
4 3/2 4 3/2 4 3/2 8 5/2
2w 1 w dw w(1 w) (1 w) dw w(1 w) (1 w)
C
3 3 3 15
3/2
8
5/2
4 1 x 1 1 x 1 1 x C
3 15
2
105. 1
dx u x u x
;
2
x 1 x 2
2u du dx
1 u
du;
2 2
tan , , sec , 1 sec
2 2
u du d u
2
1 u
2
2
2sec
sec
2
2sec 2ln sec tan 2ln 1 2ln 1
du d d C u u C x x C
106.
1/2 2
1 1 x dx;
0
2
sin , , cos , 1 cos , 0 sin 0, 1 sin
2 2 2 6
x dx d x x x
/6 /6 2
/6 /6
1 cos sin cos sin cos
0 1 cos cos d 0 cos d d 1 cos 0
lim ;
1 cos c 0 c
d
1 cos
cos , sin , sin , 2 1 cos
1 cos
u du d dv d v
1/2
/6 /6
1/2
lim 2cos 1 cos 2 1 cos sin
c 0
c c
lim 2cos
1/2 1/2 3/2
/6
4
6 1 cos 6 2cos c 1 cos c
3
1 cos
c 0
c
lim 3 1
1/2
1/2 3/2
3 4 4
3/2
2 2cos c 1 cos c
3 1 cos 6 3
1 cos c
c 0
d
1/2
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