3.3 Trigonometric Substitution

[STARTS AT 17:51]

Instructor: Let's look at number 37. This one we are not going to evaluate. We are going to replace this in terms of sine. So first notice, that we have a squared minus x squared. So we'll go ahead and make a substitution of a equals sine theta, and dx is cosine theta d theta. So the x cubed is going to be a sine cubed theta. And this is going to be 25. Oh, it should be a 5 in there. 5 sine theta, so 5 cosine theta. So that will be 125 when we cube that. Now, square root of 25 minus 25 sine squared theta. Our dx is going to become 5-- 5 cosine theta d theta. 

Well, that 25 minus-- that 25 sine squared is going to become a 5 cosine theta. 125 sine cubed theta, times 5 cosine theta d theta. Then those cancel out, and this is the integral of 125 sine cubed theta d theta. Now this says, do not evaluate. So this problem is complete. However, you'd likely want to take the sine squared in there and turn that into a cosine-- 1 minus cosine squared, and use a u substitution there. For now that is it.

[ENDS AT 19:42]