Unit 8 - Objective 5 - Odd powers of Sine and/or Cosine

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:

• sin²x + cos²x = 1
• sin²x = 1 - cos²x
• cos²x = 1 - sin²x

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