Question

Evaluate ${\int }_0^{\pi /2}si{n}^{-1}\left(cosx\right)dx$

• A. $0$
• B. $\frac{{\pi }^2}{4}$
• C. $\frac{{\pi }^2}{8}$
• D. $\frac{\pi }{2}$

Answer 1

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

We can surely evaluate this using the chain rule and using the standard integrals of inverse trigonometric function.
But lets look at the property of definite integral which will make this question a much easier one.
Using this here,
${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x).dx.$
${\int }_0^{\pi /2}si{n}^{-1}\left(cosx\right)dx={\int }_0^{\pi /2}si{n}^{-1}\left(cos\right(\frac{\pi }{2}-x\left)\right)dx$
${\int }_0^{\pi /2}si{n}^{-1}(.sinx)dx$
${\int }_0^{\pi /2}x.dx$
$=\frac{x^2}{2}{|}_{\pi /2}^0=\frac{{\pi }^2}{8}$