James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

476 Chapter 7 Techniques of Integration
SOLUTION Let
u − sin n21 x dv − sin x dx
Then du − sn 2 1d sin n22 x cos x dx v − 2cos x
so integration by parts gives
y sin n x dx − 2cos x sin n21 x 1 sn 2 1d y sin n22 x cos 2 x dx
Since cos 2 x − 1 2 sin 2 x, we have
y sin n x dx − 2cos x sin n21 x 1 sn 2 1d y sin n22 x dx 2 sn 2 1d y sin n x dx
As in Example 4, we solve this equation for the desired integral by taking the last term
on the right side to the left side. Thus we have
n y sin n x dx − 2cos x sin n21 x 1 sn 2 1d y sin n22 x dx
or y sin n x dx − 2 1 n cos x sinn21 x 1 n 2 1
n
y sin n22 x dx n
The reduction formula (7) is useful because by using it repeatedly we could eventually
express y sin n x dx in terms of y sin x dx (if n is odd) or y ssin xd 0 dx − y dx (if n is
even).
1–2 Evaluate the integral using integration by parts with the
indicated choices of u and dv.
1. y xe 2x dx; u − x, dv − e 2x dx
2. y sx ln x dx; u − ln x, dv − sx dx
3–36 Evaluate the integral.
3. y x cos 5x dx 4. y ye 0.2y dy
5. y te 23t dt 6. y sx 2 1d sin x dx
7. y sx 2 1 2xd cos x dx 8. y t 2 sin t dt
9. y cos 21 x dx 10. y ln sx dx
11. y t 4 ln t dt 12. y tan 21 2y dy
13. y t csc 2 t dt 14. y x cosh ax dx
15. y sln xd 2 dx 16. y
z
10 z dz
17. y e 2 sin 3 d 18. y e 2 cos 2 d
19. y z 3 e z dz 20. y x tan 2 x dx
21. y
xe 2x
s1 1 2xd 2 dx
22. y sarcsin xd 2 dx
23. y 1y2
x cos x dx 24. y 1
sx 2 1 1de 2x dx
0
25. y 2
y sinh y dy 26. y 2
w 2 ln w dw
0
27. y 5 ln R
1 R dR 28. y 2
t 2 sin 2t dt
2 0
29. y
x sin x cos x dx
30. y s3
arctans1yxd dx
0
0
1
1
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