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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.

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up vote 5 down vote accepted

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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I'm too slow formatting integrals, I think! – amWhy Dec 5 '12 at 20:24
Wow, that definately clears up alot. Sorry about any bad formatting when i made the question. I am new to this site. – Jeremy Rowler Dec 5 '12 at 20:30

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. +1 – Joe Johnson 126 Dec 5 '12 at 20:20

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