Mathematics - Inverse Trigonometric Function Question with Solution | TestHub

$\begin{array}{rl}& I={\int }_{1/4}^{3/4}\cos \left(2{\cot }^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)dx\right)\\ & {\int }_{1/4}^{3/4}\cos \left(2\left({\tan }^{-1}\sqrt{\frac{1+x}{1+x}}\right)\right)dx\\ & {\int }_{1/4}^{3/4}\frac{1-{\tan }^2\left({\tan }^{-1}\sqrt{\frac{1+x}{1-x}}\right)}{1+{\tan }^2({\tan }^{-1}\sqrt{\frac{1+x}{1-x}}})dx\end{array}$ $\begin{array}{rl}& ={\int }_{1/4}^{3/4}\frac{1-\left(\frac{1+x}{1-x}\right)}{1+\left(\frac{1+x}{1-x}\right)}dx={\int }_{1/4}^{3/4}\frac{-2x}{2}dx\\ & ={\int }_{1/4}^{3/4}(-x)dx=-{\left(\frac{x^2}{2}\right)}_{1/4}^{3/4}\\ & =-\frac{1}{2}\left[\frac{9}{16}-\frac{1}{16}\right]\\ & =-\frac{1}{4}\end{array}$