How to Solve Trigonometric Integrals with Odd Powers?

[physstudent1]

Homework Statement

The integral of cos[x]*(sin[x])^5dx

Homework Equations

The Attempt at a Solution

I split it into cos[x]*(sin[x])^2 * (sin[x])^2 * sin[x]

then each sin^2[x] term becomes (1-cos[x]^2)

so you end up with the integral of cos[x]*(1-cos[x]^2)^2*sin[x]dx

let u=cos[x]
-du=sin[x]dx

so it comes to the integral of -u*(1-u^2)^2 which foils out to be

-u+2u^3-u^5 then integrating this gives

-(1/2)u^2 +(2/4)*u^4 - (1/6)*u^6+c

replacing all u's with cos[x] of course at the end I just want to make sure this is correct because I have to hand it in tomarrow and this is the first of these problems I have tried and I don't want to do them all wrong.

 

[rocomath]

\int\cos x \sin^{5}xdx

Yes?

Let u=\sin xdx and it's solved.

 

[physstudent1]

I'm sorry I wasn't clear enough in the explanation you must use "the method for odd powers to evaluate the integral" since we are just learning this technique.

 

[rocomath]

I'm sorry I wasn't clear enough in the explanation you must use "the method for odd powers to evaluate the integral" since we are just learning this technique.

Well then your answer is correct!