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