Question

$\int \frac{e^x}{\sqrt{4-e^{2x}}}dx$ equals
(where $C$ is constant of integration)

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

Answer 1

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

$I=\int \frac{e^x}{\sqrt{4-e^{2x}}}dx=\int \frac{e^x}{\sqrt{2^2-(e^x{)}^2}}dx$
Let $e^x=t\Rightarrow e^{x}dx=dt$
$\therefore I=\int \frac{dt}{\sqrt{4-t^2}}=\int \frac{dt}{\sqrt{2^2-t^2}}$
$={\sin }^{-1}\left(\frac{t}{2}\right)+C={\sin }^{-1}\left(\frac{e^x}{2}\right)+C$