For \(f(x) = \int\displaystyle\frac{e^x}{\sqrt{4 - e^{2x}}}\,dx\), if the point \(\left(0, \displaystyle\frac{\pi}{2}\right)\) 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) → \(\displaystyle\frac{\pi}{3}\)
Let \(t = e^x\), so \(dt = e^x\,dx\) and \(e^{2x} = t^2\):
\( \displaystyle f(x) = \int\frac{dt}{\sqrt{4 - t^2}} \)
By the standard integral \(\int\frac{dt}{\sqrt{a^2 - t^2}} = \sin^{-1}\frac{t}{a} + C\) with a = 2:
\( \displaystyle f(x) = \sin^{-1}\frac{e^x}{2} + C \)
The point \(\left(0, \frac{\pi}{2}\right)\) lies on y = f(x), so \(f(0) = \frac{\pi}{2}\):
\( \displaystyle \sin^{-1}\frac{1}{2} + C = \frac{\pi}{2} \)
\( \displaystyle \frac{\pi}{6} + C = \frac{\pi}{2} \)
\( \displaystyle C = \frac{\pi}{3} \)