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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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Would you know what to do with $$\frac12\int \frac{1+\cos\,2x}{1+\sin\,2x} \mathrm dx$$? –  J. M. Jun 15 '12 at 10:10
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This isn't any better than what you have, but the substitution $u=\tan(x/2)$ turns any rational expression in trig functions into a rational expression in $u$. –  Gerry Myerson Jun 15 '12 at 10:46
    
Ask Wolfram, but this only gives a result (and not a good(!) explanation)... –  draks ... Jun 15 '12 at 10:48
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2 Answers 2

up vote 2 down vote accepted

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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Maybe the OP can follow all the steps you skipped, but I can't. How did you solve the first half of that split-up integral? –  Mike Jun 15 '12 at 14:18
    
Sorry :( ... i got that result from wolframalpha, I updated .. above –  Santosh Linkha Jun 15 '12 at 17:15
    
@experimentX $$\frac{\sin x}{ \sin x + \cos x} - \frac{-\cos x}{\sin x + \cos x} = 1 \neq 2$$ ;) –  qoqosz Jun 15 '12 at 19:02
    
Hmm ... thanks for correction :) –  Santosh Linkha Jun 15 '12 at 19:03
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$$\frac{1}{(1+\tan x)^{2}} = \frac{(\cos x)^{2}}{(\sin x+\cos x)^{2}}$$ $$= \frac{(\cos x)^{2}(\sin x - \cos x)^{2}}{(\cos 2x)^{2}}$$ $$= \frac{(\cos 2x - 1)(1-\sin 2x)}{2(\cos 2x)^{2}}$$ $$= 1/2 (\sec 2x)-(\sec 2x)^{2}-(\tan 2x)+(\sec 2x)(\tan 2x) $$ $$\int (\sec 2x) = 1/2\ln (\sec 2x + \tan 2x)$$ $$\int(\sec 2x)^{2}=1/2\cdot \tan 2x$$ $$\int(\tan 2x) = 1/2 \ln (\cos 2x)$$ $$\int (\sec 2x)(\tan 2x)=1/2 \sec 2x$$

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