Calculus question I'm working on a calculus problem and I need a push in the right direction. 
The problem is:
$$\int\frac{5\sin(x)}{1-5\cos(x)}dx=$$
I've played with it for hours but I can't figure out where to start. Thanks!
 A: Do a $u$-substitution with $u=1-5\cos(x)$. 
A: HINT Try a u-substitution with $u = 1 - 5\cos{x}$
A: The problem with "let $u = 1 - 5 \cos x$" is that, for beginners, it comes out of left field. Experts – even not so experts – see that substitution immediately. My own suggestion to my students was that you should check any integral against 4 kinds of known integrals: the integrals of powers $u^n$, sine, exponential, and $\sec^2$ (you have to know this one). That short list stands for more things: for powers, the special case $n = -1$; along with sine, the cosine; along with $e^u$, be prepared to consider $a^u$; in addition to $\sec^2$, $\csc^2$.
Having that list in your head, before you rush to integration by parts or general substitution, ask: is there some substitution that turns this given integral into one of those four?
In this case, the integral is certainly not $\sin u\; du$, $e^u\; du$, or $\sec^2 u\; du$. Is there any chance that it's $u^n\; du$?
If it is, then $n = -1$. that is, ask what happens if the stuff in the denominator is $u$, i.e. what happens if you set $u = 1 - 5 \cos x$. Compute $du$ and see if it collapses to $u^n\; du$. (Of course, it does.)
So if the substitution isn't obvious, this checklist gives you a suggestion of what to try.
For more detail about this approach, you could look at this post on my blog:
http://rip94550.wordpress.com/2010/01/25/calculus-organizing-techniques-of-integration/
A: What do you know (calculus-wise) about the relationship between cosine and sine?

 The derivative of the cosine function is the opposite of the sine function ($\frac{d}{dx}\cos x=-\sin x$), which suggests a $u$-substitution with $u$ having to do with cosine and $du$ having to do with sine.

