3. $\int_0^{\frac{1}{2}} \frac{x \sin ^{-1} x}{\sqrt{1-x^2}} d x=$

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Aicharya · Accepted Answer

$\int_0^{1 / 2} \frac{x \sin ^{-1} x}{\sqrt{1-x^2}} d x=\int_0^{1 / 2} x \sin ^{-1} x \cdot \frac{1}{\sqrt{1-x^2}} d x$ ...(i) Let $\sin ^{-1} x=t \Rightarrow \sin t=x$ Substituting above values in eqn. (i), we get : $\int_0^{1 / 2} \frac{x \sin ^{-1} x}{\sqrt{1-x^2}} d x=\int_0^{\pi / 6} t \sin t d t$ $\begin{aligned} & =[t(-\cos t)+\sin t]_0^{\pi / 6} \ & =\left[-\frac{\pi}{6} \cdot \cos \left(\frac{\pi}{6}ight)+\sin \left(\frac{\pi}{6}ight)-0-0ight] \ & =\left[-\frac{\pi}{6} \cdot \frac{\sqrt{3}}{2}+\frac{1}{2}ight]=\frac{1}{2}-\frac{\sqrt{3}}{12} \pi\end{aligned}$

3. $\int_0^{\frac{1}{2}} \frac{x \sin ^{-1} x}{\sqrt{1-x^2}} d x=$

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