For $f(x)=\int\frac{e^x}{\sqrt{4-e^{2x}}}dx$, if the point $\left(0,\frac{π}{2}\right)$ satisfies $y=f(x)$, then the constant of integration of the given integral is:

$\frac{π}{3}$

The correct answer is Option (2) → $\frac{π}{3}$

$f(x)=\int\frac{e^x}{\sqrt{4-e^{2x}}}dx$

let, $e^x=2\sin θ$

$e^xdx=2\cos θdθ$

$⇒dx=\frac{2\cos θ}{e^x}dθ=\frac{2\cos θ}{2\sin θ}dθ=\cot θdθ$

$∴\sqrt{4-e^{2x}}=2\cos θ$

$I=\int\frac{2\sin θ}{2\cos θ}.\cot θdθ$

$=\int dθ=θ+C$

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

Now,

$f(0)=\frac{\pi}{2}$

$⇒\sin^{-1}\left(\frac{e^0}{2}\right)+C=\frac{\pi}{2}$

$⇒C=\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}$