TITLE: Another improper integral
QUESTION [6 upvotes]: Show that :
$$\int_0^1\frac{(\sin ^{-1}x)^2}{x}\text{d}x=\frac{\pi ^2\ln 2}{4}-\frac78\zeta(3)$$
This integral is in "irresistible integrals" on page 122. I can't prove this one.

REPLY [3 votes]: Using the change of variables $ x=\cos(t) $ gives
$$\int_0^1\frac{(\sin ^{-1}x)^2}{x}\text{d}x=  \frac{1}{4}\,\int _{0}^{\pi/2 }\!{\frac { \left( \pi -2\,t \right) ^{2}\sin\left( t \right) }{\cos \left( t \right) }}{dt}.$$
Integrating the last integral using by parts techniques with $u=(\pi-2t)^2$ yields
$$  \frac{1}{4}\,\int _{0}^{\pi/2 }\!{\frac { \left( \pi -2\,t \right) ^{2}\sin
 \left( t \right) }{\cos \left( t \right) }}{dt}=-\pi\int _{0}^{\pi/2 }\ln\left(\cos \left( t \right)\right) \,dt + 2\int_{0}^{\pi/2}\,t\,\ln  \left( \cos \left( t \right)  \right) \,{dt}$$
$$  = \frac{{\pi }^{2}\ln\left( 2 \right)}{2} + 2\int_{0}^{\pi/2}\,t\,\ln  \left( \cos \left( t \right)  \right) \,{dt}.  $$
Now, I leave it here for you to finish the task. You may need to use some techniques from a previous problem.