Question

Evaluate $\int e^x\left[\frac{\sqrt{1-x^2}{\sin }^{-1}x+1}{{\sqrt{1-x}}^2}\right]dx$

Answer 1

$I=\int e^x\left[\frac{\sqrt{1-x^2}{\sin }^{-1}x+1}{\sqrt{1-x^2}}\right]dx$

$=\int e^x[{\sin }^{-1}x+\frac{1}{\sqrt{1-x^2}}]dx$

We know that $\frac{d\left({\sin }^{-1}x\right)}{dx}=\frac{1}{\sqrt{1-x^2}}$

Hence, the integral is of the form $e^x\left(f\right(x)+f^{\prime }(x\left)\right)$, where $f\left(x\right)={\sin }^{-1}x$

Now, $\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx=e^{x}f\left(x\right)+c$, where $c$ is the constant of integration.

$\therefore I=e^x{\sin }^{-1}x+c$