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

$$\begin{gathered} \mathrm{I}={\int }_0^1 \sin \left(2 {\cot }^{-1}\sqrt{\frac{1-\mathrm{x}}{1+\mathrm{x}}}\right)\mathrm{dx} \\ \sin \left(2 {\cot }^{-1}\sqrt{\frac{1-\mathrm{x}}{1+\mathrm{x}}}\right), \mathrm{put} \mathrm{x}>\mathrm{cos\theta } \\ \sin \left(2 {\cot }^{-1}\sqrt{\frac{1-\cos 2\mathrm{\theta }}{1+\cos 2\mathrm{\theta }}}\right) \\ \sin \left(2 {\cot }^{-1}(\tan \mathrm{\theta })\right) \\ =Sin\left(2 {\cot }^{-1}\left(\cot \left(\frac{\mathrm{\pi }}{2}-\mathrm{\theta }\right)\right)\right) \\ =Sin\left(\mathrm{\pi }-2\mathrm{\theta }\right)=Sin2\mathrm{\theta } \\ =\sqrt{1-{\mathrm{x}}^2} \\ \mathrm{I}={\int }_0^1 \sqrt{1-{\mathrm{x}}^2} \mathrm{dx} \\ ={\left(\frac{\mathrm{x}}{2}\sqrt{1-{\mathrm{x}}^2}+\frac{1}{2} Si{n}^{-1}\mathrm{x}\right)}_0^1 \\ =\left(0+\frac{1}{2}.\frac{\mathrm{\pi }}{2}\right)-(0+0)=\frac{\mathrm{\pi }}{4} \end{gathered}$$