Thomas Calculus 13th [Solutions]

564 Chapter 8 Techniques of Integration
2
1 2x
1 4 4 4 x
dx dw ln w ln 1
3 2
x 3 w 3 3 4
1
4
Thus the original integral is equal to
3 2
x 1 x 1 2 4 x
tan x ln 1
3 2 3 3 4
C
53. (a) u x, du dx; dv sin x dx, v cos x;
(b)
(c)
(d)
S1 x sin x dx x cos x cos x dx sin x
0 0 0
0
2 2 2
2
S2
x sin x dx x cos x cos x dx 3 sin x 3
0
3 3 3
3
S3 x sin x dx x cos x cos x dx 5 sin x 5
2 2 2
2
n 1 ( 1) 1
( 1) ( 1)
1 ( 1) n n
Sn x sin x dx ( 1) x cos x n sin x
n
n
n n
n 1 n n 1
( 1) ( n 1) ( 1) n ( 1) 0 (2n
1)
54. (a) u x, du dx; dv cos x dx, v sin x;
(b)
(c)
(d)
3 2 3 2 3 2
3
3 2
S1 x
2 cos x dx x sin x
2 2
sin x dx
2 2
cos x
2
2
5 2 5 2 5 2
5 3
5 2
S2 x
3 2 cos x dx x sin x
3 2 3 2
sin x dx
2 2
cos x
3 2
4
S3 7 2 7 2 7 2
7 5
7 2
x
5 2 x dx x sin x
5 2 5 2
x dx
2 2
x
5 2
6
S n ( 1) n (2n
1) 2 (2 1) 2 (2 1) 2
(2 1) 2 n
sin n
n
n
(2n
1) 2 (2n
1) 2
( 1) n (2n
1) (2 1) 1 (2 1) 2 1
2 ( 1) n n
2 ( 1) n
cos n
x
(2 1) 2 2
2 n
n
2 n 2 n
55.
ln 2
x
V 2 ln 2 x e dx
0
ln 2 x ln 2 x
2 ln 2 e dx 2 xe dx
0 0
x
ln 2
x
ln 2 ln 2 x
(2 ln 2) e 2 xe e dx
0 0 0
x
ln 2
2 ln 2 2 2 ln 2 e 2 ln 2 2
0
2 (1 ln 2)
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