Mathematics

# Solve the integral $\int \sqrt{\dfrac{1+x}{1-x}}dx$.

##### SOLUTION

Put $x=\cos 2\theta$ on differentiating with respect to $\theta$ , and we get

$dx=-2\sin 2\theta d\theta$

put given function  and solve given function

$\int{\sqrt{\dfrac{1-x}{1+x}}}dx$

$=\int{\sqrt{\dfrac{1+\cos 2\theta }{1-\cos 2\theta }}\left( -2\sin 2\theta d\theta \right)}$

$=-2\int{\sqrt{\dfrac{1+\left( 2{{\cos }^{2}}\theta -1 \right)}{1-\left( 1-2{{\sin }^{2}}\theta \right)}}\sin 2\theta d\theta }$

$=-2\int{\sqrt{\dfrac{2{{\cos }^{2}}\theta }{2si{{n}^{2}}\theta }}}\sin 2\theta d\theta$

$=-2\int{\sqrt{{{\cot }^{2}}\theta }}\sin 2\theta d\theta$

$=-2\int{\cot \theta }\sin 2\theta d\theta$

$=-2\int{\dfrac{\cos \theta }{\sin \theta }2\sin \theta \cos \theta d\theta }$

$=-4\int{{{\cos }^{2}}\theta d\theta }$

$=-4\int{\left( \dfrac{1+\cos 2\theta }{2} \right)}d\theta$

$=-2\int{1d\theta -2\int{\cos 2\theta d\theta }}$

On integration, we get

$=-2\theta -2\dfrac{\sin 2\theta }{2}+C$

$=-2\theta -\sin 2\theta +C$       By equation (1),     $\theta =\dfrac{1}{2}{{\cos }^{-1}}x$

Put here and we get,

$=-2\times \dfrac{1}{2}{{\cos }^{-1}}x-\sin 2\left( \dfrac{1}{2}{{\cos }^{-1}}x \right)+C$

$=-{{\cos }^{-1}}x-\sin {{\cos }^{-1}}x+C$

$=-{{\cos }^{-1}}x-\sin {{\sin }^{-1}}\sqrt{1-{{x}^{2}}}+C$

$=-{{\cos }^{-1}}x-\sqrt{1-{{x}^{2}}}+C$

$=-\sqrt{1-{{x}^{2}}}-{{\cos }^{-1}}x+C$

It is complete solution.

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Subjective Medium Published on 17th 09, 2020
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