Question

${\int }_0^1\frac{{\sin }^{-1}x}{x}dx=\frac{\pi }{k}\log k$. Find the value of $k$.

Answer 1

Put $x=\sin \theta$ and adjust the limits
$I={\int }_0^{\pi /2}\frac{\theta }{\sin \theta }\cos \theta d\theta ={\int }_0^{\pi /2}\theta \cot \theta d\theta .$
Integrate by parts
$I={\left[\theta \log \sin \theta \right]}_0^{\pi /2}-{\int }_0^{\pi /2}\log \sin \theta .1d\theta$
$=\left(\frac{\pi }{2}.0\right)-\underset{\theta \to 0}{lim}\left(\theta \log \sin \theta \right)-\frac{\pi }{2}\log \frac{1}{2}$
$=0+\frac{\pi }{2}\log 2$

$\therefore k=2$