Question

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

Answer 1

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

Let $x=\sin tdx=\cos tdt$

$I={\int }_0^{\frac{\pi }{4}}t\cos tdt={t\sin t}_0^{\frac{\pi }{4}}-{\int }_0^{\frac{\pi }{4}}1.\sin tdt={t\sin t+\cos t}_0^{\frac{\pi }{4}}=\left[\right(\frac{\pi }{4}\sin \frac{\pi }{4}+\cos \frac{\pi }{4})-(0+\cos 0\left)\right]=[\frac{\pi }{4\sqrt{2}}+\frac{1}{\sqrt{2}}-1]=\frac{\pi +4-4\sqrt{2}}{4\sqrt{2}}$