Evaluate $$\int_{0}^{1} \frac{\sin^{-1} x}{\sqrt{1-x^2}} dx$$
Step-by-Step Solution
Key Concept: General
Solution:<br>$\sin^{-1} x = t$<br>$\frac{dx}{\sqrt{1-x^2}} = dt$<br>at $x = 0, t = 0$<br>at $x = 1, t = \pi/2$<br>$$\int_{0}^{\pi/2} t dt = \left( \frac{t^2}{2} \right)_0^{\pi/2} = \frac{\pi^2}{8}$$
Correct Answer: \frac{\pi^2}{8}