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The question is to find

$\displaystyle \int \frac {\sin (2x)}{1+\cos^2x}.$

Can anyone help me? I need all the steps, because I need to understand what to do. Thank you.

3 Answers 3

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You can use the identity: $ \sin(2x)=2\sin(x)\cos(x). $ Then use a $u$-substitution with $u=1+\cos^2(x)$.

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    Wow, that definately clears up alot. Sorry about any bad formatting when i made the question. I am new to this site.2012-12-05
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By the double angle formula, $\sin(2x) = 2 \sin(x)\cos(x)$

$\int \frac{\sin(2x)}{1 + \cos^2(x)}dx = \int \frac{2\sin(x)\cos(x)}{1+\cos^2(x)}dx$

Let $u = 1 + \cos^2(x)$ $du = -2\sin(x)\cos(x) dx$

so...substituting, we get: $\int \frac{2\sin(x)\cos(x)}{1+\cos^2(x)}dx = \int -\frac{1}{u} du$

Can you take it from here?

Integrate with respect to $u$, then "back" substitute $u = 1 + \cos^2(x)$ into the result.

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Note that $\sin(2x)=2\sin(x)\cos(x)$

Therefore the problem reduces to finding the integral: $\int \frac {2\sin(x)\cos(x)}{1+\cos^2(x)}dx=-\log(1+\cos^2(x))+C$

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    A photo finish. But, you beat me by a second. +12012-12-05