Question

Evaluate the following integrals:

$\int \cos \left\{2{\cot }^{-1}\sqrt{\frac{1+x}{1-x}}\right\}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \cos \left\{2{\cot }^{-1}\sqrt{\frac{1+x}{1-x}}\right\}dx \\ \mathrm{Let}x=\cos 2\theta \\ \mathrm{On}\mathrm{differentiating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \mathrm{d}x=-2\sin 2\theta \mathrm{d}\theta \\ \therefore I=-\int \cos \left\{2{\cot }^{-1}\sqrt{\frac{1+\cos 2\theta }{1-\cos 2\theta }}\right\}2\sin 2\theta \mathrm{d}\theta \\ =-2\int \cos \left\{2{\cot }^{-1}\sqrt{\frac{2{\cos }^2\theta }{2{\sin }^2\theta }}\right\}\sin 2\theta \mathrm{d}\theta \\ =-2\int \cos \left\{2{\cot }^{-1}\left(\cot \theta \right)\right\}\sin 2\theta \mathrm{d}\theta \\ =-2\int \cos 2\theta \sin 2\theta \mathrm{d}\theta \\ =-\int \sin 4\theta \mathrm{d}\theta \\ =\frac{\cos 4\theta }{4}+c_1 \\ =\frac{1}{4}\left(2{\cos }^{2}2\theta -1\right)+c_1 \\ =\frac{1}{2}{x}^2-\frac{1}{4}+c_1 \\ =\frac{1}{2}{x}^2+c,\mathrm{where}c=-\frac{1}{4}+c_1 \\ \mathrm{Hence},\int \cos \left\{2{\cot }^{-1}\sqrt{\frac{1+x}{1-x}}\right\}dx=\frac{1}{2}{x}^2+c \end{gathered}$$