Please Integrate the following definite integral:

$\Rightarrow {\int }_{-1}^1 \frac{x}{\sqrt{1 - x^2}}, {\sin }^{-1} \left(2x \sqrt{1 - x^2}\right). dx$

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{I}={\int }_{-1}^1\frac{\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}{\sin }^{-1}\left(2\mathrm{x}\sqrt{1-{\mathrm{x}}^2}\right) \\ \mathrm{Let} \mathrm{x}=\mathrm{sint} \\ \mathrm{as} \mathrm{x} \mathrm{goes} \mathrm{from} -1 \mathrm{to} 1, \mathrm{hence} \mathrm{t} \mathrm{goes} \mathrm{from} -\frac{\mathrm{\pi }}{2} \mathrm{to} \frac{\mathrm{\pi }}{2} \\ \mathrm{I}={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}} \frac{\mathrm{sint}}{\sqrt{1-{\sin }^2\mathrm{t}}}{\sin }^{-1}\left(2\mathrm{sint}\sqrt{1-{\sin }^2\mathrm{t}}\right) \\ ={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}} \frac{\mathrm{sint}}{\sqrt{1-{\sin }^2\mathrm{t}}}{\sin }^{-1}\left(2\mathrm{sint}\sqrt{1-{\sin }^2\mathrm{t}}\right) \\ ={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}} \frac{\mathrm{sint}}{\mathrm{cost}}{\sin }^{-1}\left(2\mathrm{sintcost}\right) \\ ={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}} \mathrm{tant} {\sin }^{-1}\left(\sin 2\mathrm{t}\right) \\ =2{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}} \mathrm{t}.\mathrm{tant} \\ \mathrm{The} \mathrm{integration} \mathrm{of} \mathrm{t}.\mathrm{tant} \mathrm{is} \mathrm{not} \mathrm{possible} \mathrm{because} \mathrm{t}.\mathrm{tant} \mathrm{is} \mathrm{not} \mathrm{integrable} \\ \mathrm{in} \mathrm{terms} \mathrm{of} \mathrm{elementary} \mathrm{function} \mathrm{which} \mathrm{are} \mathrm{taught} \mathrm{to} \mathrm{us}. \end{gathered}$$

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