I have a hard time solving integrals of trig equations. Any advice? I'm working through a section of integrals of trig functions, and I always have the hardest time knowing off the bat the next step to make.  I know the techniques, but knowing which step to take is difficult.  There's a tree of possible solutions and I'm expected to know the correct route, and I don't.  I'm magically supposed to know that the half-angle formula is the correct approach, or whatever it is.
It's a real sticking point right now.  How do you approach these problems?  Any advice?
Btw, the question that really threw me was this:
$\int cos^4(2t)dt$
The solution required two shots of half-angle formula, and a substitution.  I didn't magically know that. @_@
 A: I would start by exhausting powers of sine and cosine. Odd powers are easy since you can always make the $u$-sub. of either sine or cosine, then $\sin^2(\theta)+\cos^2(\theta)=1$ allows you to massage the expression as needed. Then I would go to even powers of secant and tangent, still pretty easy as long as you know the $\tan^2(\theta)+1=\sec^2(\theta)$. Past that, I look for having an odd number of sine and cosine factors, often a $u=\sin(\theta)$ or $u=\cos(\theta)$ substitution works. For example, $\int \sin^3(x)\cos^2(x) dx$. However, for $\int \frac{dx}{\cos^3(x)}$ thought is required (at least for me). 
For many students, inability to recall trig. identities is a big hole in your armor. One way to help fill it is to learn how imaginary exponentials can be used to derive trig. identities. For example, $ \int \sin(3x)\cos(2x) dx $ is quite mysterious until you know:
$$ \sin(3x)\cos(2x) = \frac{1}{2i}\biggl[e^{3ix}-e^{-3ix}\biggr]\frac{1}{2}\biggl[e^{2ix}+e^{-2ix}\biggr] = \underbrace{\frac{1}{4i}\biggl[e^{5ix}-e^{-5ix}\biggr]}_{\frac{1}{2}\sin(5x)}+\underbrace{\frac{1}{4i}\biggl[e^{ix}-e^{-ix}\biggr]}_{\frac{1}{2}\sin(x)}$$
Hence $\int \sin(3x)\cos(2x)dx = -\frac{-1}{10}\cos(5x)-\frac{1}{2}\cos(x)+C$. Of course, you can memorize the appropriate trig. identities to do these sort of problems, but I find it comforting to know on an Island with no reference texts or stack exchange I can still derive all the trig identies my heart desires.
Beyond this, things like $\sec(x)$ need a trick for speedy solution $u = \sec(x)+\tan(x)$ and common sense indicates $\csc(x)$ is similar. 
All I'm telling you, and perhaps this is not what you want to hear, is to practice. But, try to look for patterns as you do problems. Ask yourself, if this problem had been twisted a bit could I still make this approach work... if you're short on time maybe use wolfram alpha to check if a conjecture is in the right direction or not.
A: I believe knowing how to do it is good enough. And actually not being able to (or to see the proper route to) evaluate the integral instantly isn't such a problem. You're not going to get the points for not solving it, but that's not important. When you see $\int \tan^2(x) \sec^4(x) \, \ dx$, you should obviously not be thinking about using the partial fraction method, say, to crack it. All it takes is a simple substitution and that's what important is if you can narrow down the potential ways to evaluate it and apply the one that works, you're good.
A: Its just a list of substitutions and more importantly ; when to use them.
you can always reduce the powers and that is important.
So when you see e.g. $\sin^4(x)$ , you know you could maybe rewrite to $\left(\frac {1-\cos(2x)}{2}\right)^2$.
for high powers this often works better than partial or substitute.
There are no more than about 9 tricks. Practice.
