Thanks for the help. I ran into another brick wall though. I worked the problem down to the part where I get this:
[f(x)]^2=\int{\sin 2\theta ( \cos^6 2 \theta-cos^8 2\theta) d\theta
And substitute:
u: u = \cos 2\theta
du: du = -2\sin 2\theta
du: -\frac{1}{2}du = \sin 2\theta
So...
Hey y'all. I'm new to the forum, and have a problem that I've been working on all night long. I'm having issues previewing the Latex, so bear with me. I'll post the work I've done so far if the problem code shows up. Thanks.
\int{\sin^{\frac{3}{2}}2\theta\cos^{3}2\theta} d\theta
Now, the...