Question

$\int (e^{{\sin }^{-1}x}+\frac{1}{\sqrt{1-x^2}})dx$

• A. $e^{{\sin }^{-1}x}\left(\frac{\sqrt{1-x^2}+x}{2}\right)+{\sin }^{-1}x+c$
• B. $e^x(1+{\tan }^{2}x)+c$
• C. $e^x\tan 2x+c$
• D. $e^x(1+\tan x)+c$

Answer 1

The correct option is A $e^{{\sin }^{-1}x}\left(\frac{\sqrt{1-x^2}+x}{2}\right)+{\sin }^{-1}x+c$

Let $I=\int e^{{\sin }^{-1}x}dx$ and $J=\int \frac{1}{\sqrt{1-x^2}}dx$

Let $x=\sin \theta$

$\therefore dx=\cos \theta d\theta$

$\therefore I=\int e^{\theta }\cos \theta d\theta$

Using chain rule of integration,

$I=\cos \theta \int e^{\theta }d\theta -\int (-\sin \theta )e^{\theta }d\theta +c$

$\therefore I=e^{\theta }\cos \theta +\sin \theta \int e^{\theta }d\theta -\int \cos \theta e^{\theta }d\theta +c$

$\therefore I=e^{\theta }\cos \theta +e^{\theta }\sin \theta -I+c$

$\therefore I=e^{\theta }\left(\frac{\cos \theta +\sin \theta }{2}\right)+c$

Substituting $\theta ={\sin }^{-1}x$, we have

$I=e^{{\sin }^{-1}x}\left(\frac{\sqrt{1-x^2}+x}{2}\right)+c$

Now, for $J$ also, assume that $x=\sin \theta$

$\therefore J=\int \frac{1}{\sqrt{1-{\sin }^2\theta }}\cos \theta d\theta$

$\therefore J=\int d\theta$

$\therefore J=\theta +c$

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

Hence, required answer is $I=e^{{\sin }^{-1}x}\left(\frac{\sqrt{1-x^2}+x}{2}\right)+{\sin }^{-1}x+c$