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:
Answer & explanation
Correct answer: option 2
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,
$f(0)=\frac{\pi}{2}$
$⇒\sin^{-1}\left(\frac{e^0}{2}\right)+C=\frac{\pi}{2}$
$⇒C=\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}$