How do I solve $\int\frac{\cos^2(x)}{\sin(x)}\ dx$ without using Weierstass Substitution? Every problem that I've put into wolfram alpha lately gives me instructions to substitute $\tan(\frac{x}{2})$, but I haven't been taught how to do that, nor can I understand how it works anyhow...
 A: $$ \int \frac{\cos^2(x)}{\sin(x)} \; \mathrm dx$$
You should know the Pythagorean Identity:
$$ \sin^2(x) + \cos^2(x) = 1 $$
So:
$$ \int \frac{1 - \sin^2(x)}{\sin(x)} \; \mathrm dx$$
$$ \int \left(\frac{1}{\sin(x)} - \sin(x) \right) \; \mathrm dx$$
$$ \int \left(\csc(x) - \sin(x)\right) \; \mathrm dx $$
$$ \int \csc(x) \; \mathrm dx - \int \sin(x) \; \mathrm dx $$

There is a nice little explanation of the Weierstrass Substitution here. Basically, any function using $\sin$ and $\cos$ can be expressed as rational functions of $\tan(x/2)$, so you can then easily integrate most integrals by method of partial fractions (the paper linked goes further than that).
A: Hint:
write this integral as $$ \int \frac{1 - \sin^2(x)}{\sin(x)} \; \mathrm dx$$
A: $$ \int \frac{\cos^2(x)}{\sin(x)} \mathrm dx = \int \csc(x) \; \mathrm dx  - \int \sin(x) \; \mathrm dx $$
A: Here's one approach.
First rewrite the integral as
$$
\int \frac{\cos^2(x)}{\sin(x)} dx
= \int \frac{\cos^2(x)}{\sin^2(x)} \sin(x) dx
= \int \frac{\cos^2(x)}{1-\cos^2(x)} \sin(x) dx .
$$
Using the substitution $u = \cos(x)$, this becomes
$$
- \int \frac{u^2}{1-u^2} du ,
$$
which can now be computed by using partial fractions.
I'll leave the rest of the details to you.
