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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...

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    $\begingroup$ This falls under the category $\cos^m x\sin^n x$ where $m$ and $n$ are integers, not necessarily positive. Say $m$ is odd. Then some manipulation and the substitution $u=\sin x$ lead to the integral of a rational function. $\endgroup$ – André Nicolas Mar 10 '13 at 21:36
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    $\begingroup$ see en.wikipedia.org/wiki/Weierstrass_substitution which really is a helpful thing to know. $\endgroup$ – Will Jagy Mar 10 '13 at 21:38
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$$ \int \frac{\cos^2(x)}{\sin(x)} \; \mathrm dx$$

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).

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  • $\begingroup$ Indeed I should have known that... I'm starting to think that my brain is going numb from all the trig integrals I've done. Thanks for the insight. $\endgroup$ – agent154 Mar 10 '13 at 21:28
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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.

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Hint:

write this integral as $$ \int \frac{1 - \sin^2(x)}{\sin(x)} \; \mathrm dx$$

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$$ \int \frac{\cos^2(x)}{\sin(x)} \mathrm dx = \int \csc(x) \; \mathrm dx - \int \sin(x) \; \mathrm dx $$

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