Correct Answer - Option 3 :
\(\rm \dfrac{\pi^2}{8}\)
Concept:
\(\rm \displaystyle\int x^n \;dx = \dfrac{x^{n+1}}{n+1} + c\)
\(\rm {d(sin^{-1}x)\over dx} = {1\over {\sqrt {1\ -\ x^2}}}\)
Calculation:
Given: \(\rm \displaystyle\int_0^1 \dfrac{\sin^{-1} \ x}{\sqrt {{1-x^2}}} \ dx\)
Let sin-1 x = t
\(\Rightarrow \rm \frac{dx}{\sqrt {1-x^2}} = dt\)
Now,
\(\rm \rm \displaystyle\int_0^1 \dfrac{sin^{-1} \ x}{\sqrt {{1-x^2}}} \ dx = \rm \displaystyle\int_{0}^{\pi /2} t\;dt\)
\(\rm = \left[\dfrac{t^2}{2}\right]_{0}^{\pi/2} \\ = \dfrac{\pi^2}{8}\)