Explanation

(c)
Given, $I_{1} = \int_{}^{}{\sin^{- 1}{x\ dx}}$
and $I_{2} = \int_{}^{}{\sin^{- 1}\sqrt{1 - x^{2}}}\ dx$
$\Rightarrow I_{2} = \int_{}^{}{\cos^{- 1}x}\ dx$
Now, $I_{1} + I_{2} = \int_{}^{}{(\sin^{- 1}x} + \cos^{- 1}x)dx$
$= \int_{}^{}\frac{\pi}{2}dx = \frac{\pi}{2}\ x$
$\therefore\ I_{1} + I_{2} = \frac{\pi}{2}\ x$