Question

Integrate the function: $\frac{\sin^{-1}x}{\sqrt{1-x^2}}$

Answer 1

Let sin-1 x = t

∴ $\frac{1}{\sqrt {1-x^2}}dx=dt$

$\Rightarrow \int\frac{\sin^{-1}x}{\sqrt {1-x^2}}dx=\int t dt$

= $\frac{t^2}{2}+C$

= $\frac{(sin^{-1} x)^2}{2}+C$