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I can solve this integral $ \int\frac{1}{(1+\tan x)^2} dx $ using the substitution $t=\tan x$ i.e $x=\arctan t$. Does anyone know another way to solve this integral?

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    Ask [Wolfram](http://tinyurl.com/cfco8v8), but this only gives a result (and not a good(!) explanation)...2012-06-15

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Using J.M.'s suggestion $ \int \frac{1}{( 1 + \tan x)^2} dx = \int \frac{\frac 12 (1 + \cos 2x)}{ \sin ^2x +2 \sin x \cos 2 + \cos ^2 x} dx = \frac 12 \int \frac{1 + \cos 2x}{1 + \sin 2x}dx$ $ = \frac 12 \left [ \int \frac 1 {1 + \sin 2x} dx + \int \frac{\cos 2x}{1 + \sin 2x}dx\right ]$

$ = \frac 1 2 \frac{\sin x}{\sin x + \cos x} + \frac 1 4 \ln (1 + \sin 2x) + C$

$ = \frac 12 \left [ \frac{-1}{1 + \tan x} + \ln(\sin x + \cos x)\right ] + C $

EDIT:: first half of split up integral $ \int \frac{1}{1 + \sin 2x} dx = \frac 1 2 \int \frac{1}{1 + \sin u} du $ and Weierstrass substitution $ \frac 12 \int \frac{1}{1 + \frac{2t}{1 + t^2}}\frac{2dt}{1 + t^2} = \int \frac{1}{(1 + t)^2} dt $ $ \implies \frac{-1}{1 + t} = \frac{-1}{1 + \tan x} = \frac{-\cos x}{\sin x + \cos x}$

From wolfram I got $ \int \frac{\sin x}{ \sin x + \cos x} $ Using Weierstrass substitution I got $ \frac{-\cos x}{\sin x + \cos x} $ And it seems $ \frac{\sin x}{ \sin x + \cos x} - \frac{-\cos x}{\sin x + \cos x} = 1 $ Hence both are vaid :D :D

And by clicking show steps at Wolframalpha enter image description here

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    Hmm ... thanks for correction :)2012-06-15
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$\frac{1}{(1+\tan x)^{2}} = \frac{(\cos x)^{2}}{(\sin x+\cos x)^{2}}$ $= \frac{(\cos x)^{2}(\sin x - \cos x)^{2}}{(\cos 2x)^{2}}$ $= \frac{(\cos 2x - 1)(1-\sin 2x)}{2(\cos 2x)^{2}}$ $= 1/2 (\sec 2x)-(\sec 2x)^{2}-(\tan 2x)+(\sec 2x)(\tan 2x) $ $\int (\sec 2x) = 1/2\ln (\sec 2x + \tan 2x)$ $\int(\sec 2x)^{2}=1/2\cdot \tan 2x$ $\int(\tan 2x) = 1/2 \ln (\cos 2x)$ $\int (\sec 2x)(\tan 2x)=1/2 \sec 2x$