Integrating functions of sin to a power

[howsockgothap]

Homework Statement

Integrate sin⁵(x/3)dx on the interval [0,pi]

Homework Equations

sin²(x)=1-cos²(x)
dcos(x)dx=-sin(x)

The Attempt at a Solution

I split off one of the sins and then set my integral equal to (sin²(x/3))²*sin(x/3) switching in 1-cos²(x/3) afterward. Then I set u = cos (x/3) and factored that in. I also changed my interval at this point figuring when x=pi, u=1/2 and when x=0, u=1.

So I then had -3∫(1-u²)²du on the interval [1/2,1]

I worked it out and got -3cos(1/3)+2cos³(1/3)-(3/5)cos⁵(1/3)+3cos(1/6)-2cos³(1/6)+(3/5)cos⁵(1/6)

Given how long winded that answer is-- my answer has to be submitted in exact numbers-- it doesn't seem right to me. What have I done wrong?

 

[Dick]

Why don't you just work out the integral of 3*(1-u^2)^2 on [1/2,1]?? You don't need any cosines to do that. The last expression looks wrong anyway, I'd only expect to see cos(pi/3) and cos(0) in there.

 

[lurflurf]

You do not need to change variables back to cos(x), which you did wrong as we had the interval [0,pi] we should have (if we needlessly change back) cos(pi/3) type things, which can be evaluated, not cos(1/3) type things that cannot.

 

[howsockgothap]

Right, of course. thanks!