In the first section of this unit, you integrated powers of sine as long as there was a single power of cosine to be the du.
When the integral contains only an odd power of sine (or cosine) then use the trig identities:
Peel off and save one sine (or cosine) to be the "du" and replace the rest.
Problem 1:
sin³x dx | = sin²x sin x dx | save sin x |
= (1 - cos²x) sin x dx | multiply across | |
= (sin x - cos²x sin x) dx | integrate one term at a time | |
= sin x dx - cos²x sin x dx | ||
ok | u = cos x, du = -sin x dx, "fix" the -1 | |
= sin x dx - (-1) cos²x (-1) sin x dx | ||
= -cos x - (-1) + C | ||
= -cos x + (1/3) cos³x + C |
Problem 2:
cos³(5t) dt | = (cos²5t) cos 5t dt = (1 - sin²5t) cos 5t dt |
= cos 5t dt - sin²5t cos 5t dt | |
= | |
= (1/5) sin 5t - (1/5) + C | |
= (1/5) sin 5t - (1/15) sin³5t + C |