$\underset{0}{\overset{1}{\int }}\sin \left(2{\tan }^{-1}\sqrt{\frac{{\displaystyle 1+x}}{{\displaystyle 1-x}}}\right)dx=$

$$\begin{gathered} \underset{0}{\overset{1}{\int }}\sin \left(2{\tan }^{-1}\sqrt{\frac{{\displaystyle 1+x}}{{\displaystyle 1-x}}}\right)dx \\ put x=\cos 2\theta \\ then \sin \left[2{\tan }^{-1}\sqrt{\frac{1+x}{1-x}}\right] \\ =\sin \left[2{\tan }^{-1}\sqrt{\frac{1+\cos 2\theta }{1-\cos 2\theta }}\right] \\ =\sin \left[2{\tan }^{-1}\left(cot\theta \right)\right] \\ =\sin \left[2{\tan }^{-1}\left(\tan \left(\frac{\pi }{2}-\theta \right)\right)\right] \\ =\sin \left[2\left(\frac{\pi }{2}-2\theta \right)\right] \\ =\sin 2\theta \\ =\sqrt{1-{\cos }^{2}2\theta }=\sqrt{1-x^2} \end{gathered}$$

$$\begin{gathered} I=\underset{0}{\overset{1}{\int }}\sqrt{1-x^2}dx \\ ={\left[\frac{x}{2}\sqrt{1-x^2}\right]}_0^1+\frac{1}{2}{\left[{\sin }^{-1}\frac{x}{1}\right]}_0^1 \\ =(0-0)+\frac{1}{2}\left(\pi /2\right) \\ =\pi /4 \end{gathered}$$

$\left[\underset{ }{\overset{ }{Formula\int \sqrt{a^2-x^2}}}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+c\right]$