Evaluate: \(\displaystyle\int e^{\sin^{-1}x}\left(\dfrac{x + \sqrt{1 - x^2}}{\sqrt{1 - x^2}}\right)dx\)

Answer: \(x\,e^{\sin^{-1}x} + c\)

Put \(\sin^{-1}x = t\), so \(x = \sin t\), \(dx = \cos t\,dt\) and \(\sqrt{1 - x^2} = \cos t\).

\(I = \displaystyle\int e^t\left(\dfrac{\sin t + \cos t}{\cos t}\right)\cos t\,dt = \int e^t(\sin t + \cos t)\,dt\).

This is of the form \(\int e^t[f(t) + f'(t)]\,dt = e^tf(t) + c\) with \(f(t) = \sin t\), \(f'(t) = \cos t\).

\(I = e^t\sin t + c = x\,e^{\sin^{-1}x} + c\).

(Check: \(\dfrac{d}{dx}\left(xe^{\sin^{-1}x}\right) = e^{\sin^{-1}x} + \dfrac{x\,e^{\sin^{-1}x}}{\sqrt{1 - x^2}} = e^{\sin^{-1}x}\dfrac{x + \sqrt{1 - x^2}}{\sqrt{1 - x^2}}\).)