Question

Integrate $\int \sqrt{\frac{1+x}{1-x}}dx,on(-1,1).$

Answer 1

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

Put $x=\cos 2\theta$

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

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

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

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

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

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

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

$=-{{\left[{\cos }^{-1}x+\sqrt{1-x^2}\right]}_{-1}}^{{}_1}$

$=\left[\left({\cos }^{-1}1+\pi \right)-\left({\cos }^{-1}-1+0\right)\right]$

$=\left[0-\pi \right]$

$=\pi$