# Integral of a trigonometric function [duplicate]

Possible Duplicate:
Evaluating $\int P(\sin x, \cos x) \text{d}x$

How do I integrate the following function? $$\frac{\sin 2x}{(1 + \cos^2x)^2}?$$

Thanks.

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## marked as duplicate by Aryabhata, tomasz, t.b., Asaf Karagila, J. M.Aug 19 '12 at 3:58

Let $u = 1+\cos^2 x$. From here:

$du=-2\cos x \sin x dx= -\sin 2x dx$

Therefore:

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

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You have to make a substitution: $$u=cos^2(x)$$ Then you obtaine the integral: $$- \int \frac {1}{(1+u)^2}du$$ Now make a substitution:$s=u+1$ and then you have the result: $$\int \frac{sin(2x)}{(1+cos^2(x))^2}=\frac{2}{cos(2x)+3}$$

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Use $\sin2x = 2 \sin x \, \cos x$ and substitute $\varphi = \cos x$.

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Should be $\sin(2x) = 2\sin(x)\cos(x)$. – Patrick Mar 23 '12 at 15:11