Question

Evaluate $\int \frac{e^x}{\sqrt{4-e^{2x}}}dx.$. The solution is
${\sin }^{-1}\left(\frac{e^x}{k}\right)+C$ Find $k$.

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

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