Question

${\int }_0^{\frac{1}{2}}\frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}dx$

Answer 1

Let I = ${\int }_0^{\frac{1}{2}}\frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}dx$

Put $x=\sin \theta$

$\therefore dx=\cos \theta d\theta$

When $x\to 0,\theta \to 0$

When $x\to \frac{1}{2},\theta \to \frac{\mathrm{\pi }}{6}$

$$\begin{gathered} \therefore I={\int }_0^{\frac{\mathrm{\pi }}{6}}\frac{\sin \theta {\sin }^{-1}\left(\sin \theta \right)}{\cos \theta }\cos \theta d\theta \\ ={\int }_0^{\frac{\mathrm{\pi }}{6}}\theta \sin \theta d\theta \end{gathered}$$

Applying integration by parts, we have

$$\begin{gathered} I={\overline{)\theta \left(-\cos \theta \right)}}_0^{\frac{\mathrm{\pi }}{6}}-{\int }_0^{\frac{\mathrm{\pi }}{6}}1\times \left(-\cos \theta \right)d\theta \\ =-\left(\frac{\mathrm{\pi }}{6}\cos \frac{\mathrm{\pi }}{6}-0\right)+{\int }_0^{\frac{\mathrm{\pi }}{6}}\cos \theta d\theta \\ =-\frac{\mathrm{\pi }}{6}\times \frac{\sqrt{3}}{2}+{\overline{)\sin \theta }}_0^{\frac{\mathrm{\pi }}{6}} \\ =-\frac{\mathrm{\pi }}{4\sqrt{3}}+\left(\sin \frac{\mathrm{\pi }}{6}-\sin 0\right) \\ =-\frac{\mathrm{\pi }}{4\sqrt{3}}+\left(\frac{1}{2}-0\right) \\ =\frac{1}{2}-\frac{\mathrm{\pi }}{4\sqrt{3}} \end{gathered}$$