Question

Find the integration of $\int \frac{e^{{\sin }^{-1}x}}{\sqrt{1-x^2}}.dx$

• A. $e^{{\sin }^{-1}x}+C$
• B. $e^{-{\sin }^{-1}x}+C$
• C. $\frac{e^{{\sin }^{-1}x}}{2}+C$
• D. $e^{\sin x}+C$

Answer 1

The correct option is A $e^{{\sin }^{-1}x}+C$

Given, $\int \frac{e^{{\sin }^{-1}x}}{\sqrt{1-x^2}}.dx$

Let $t={\sin }^{-1}x$
$⟹dt=\frac{1}{\sqrt{1-x^2}}.dx$
$\Rightarrow dx=\left(\sqrt{1-x^2}\right).dt$
Putting these value in above integration
$\therefore$ we have $I=\int e^t.dt=e^t+C$
$\therefore$ $I=e^{{\sin }^{-1}x}+C$