Question

${\int }_0^{\pi /2}{\sin }^{-1}\left(\cos x\right)dx=$

• A. $\frac{{\pi }^2}{4}$
• B. $\frac{{\pi }^2}{2}$
• C. $\frac{{\pi }^2}{8}$

Answer 1

The correct option is C $\frac{{\pi }^2}{8}$

${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x).dx.$
${\int }_0^{\pi /2}{\sin }^{-1}\left(\cos x\right)dx={\int }_0^{\pi /2}{\sin }^{-1}\left(\cos \right(\frac{\pi }{2}-x\left)\right)dx$
${\int }_0^{\pi /2}{\sin }^{-1}\left(\sin x\right)dx$
${\int }_0^{\pi /2}x.dx$
$=\frac{x^2}{2}{|}_0^{\pi /2}=\frac{{\pi }^2}{8}$