3.2 Trigonometric Integrals

[STARTS AT 5:37]

Instructor: Number 15, I'll do something very similar with this. I've got a cosine cubed, so I'm going to pull off two of those. So this is going to be cosine squared. Well, that's going to be 1 minus sine squared. So I'm just going to go ahead and write that. 1 minus sine squared x times cosine x, sine squared x. 

Now I have a single cosine and I have actually all even powers of sine. So this, rewriting this, I'm going to distribute the sine this way. So that is a sine squared x minus sine to the fourth x, and all that's times cosine, x, dx. My u substitution at this point will be u equals sine. Sine x, du is cosine x, dx, so that this is the integral of u squared minus u to the fourth, du. 

All right, so using my power rule, u cubed over 3, minus u to the fifth over 5, plus C, which substituting u back in, sine cubed over 3, minus sine to the fifth x over 5, plus c. And you might notice some similarities between these two problems, one with sine, one with cosine. The signs is the only thing that's different-- no pun intended, really. The positive and negativeness of each one of those is the only difference there.

[ENDS AT 7:30]