Question

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

Answer 1

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

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

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

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

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

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

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

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

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

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

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

On integration, we get

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

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

Put here and we get,

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

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

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

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

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

It is complete solution.