Question

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

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{3}$
• C. $\frac{\pi}{6}$
• D. 0

Answer 1

The Correct Option is A

The integral is:

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

We simplify the denominator:

\[\sqrt{4-e^{2x}}=\sqrt{(2)^{2}-(e^{x})^{2}}\]

This suggests a substitution of the form \(e^{x}=2~sin~\theta\), where \(e^{2x}=4~sin^{2}\theta\). Then:

\[dx=\frac{2~cos~\theta}{e^{x}}d\theta=\frac{2~cos~\theta}{2~sin~\theta}d\theta=cot~\theta~d\theta\]

Substituting into the integral:

\[f(x)=\int\frac{2~sin~\theta}{\sqrt{4-4~sin^{2}\theta}}cot~\theta~d\theta\]

Using \(\sqrt{4-4~sin^{2}\theta}=2~cos~\theta\):

\[f(x)=\int\frac{2~sin~\theta}{2~cos~\theta}cot~\theta~d\theta=\int d\theta\]

Thus:

\[f(x)=\theta+C\]

Returning to \(e^{x}=2~sin~\theta\), we have \(sin~\theta=\frac{e^{x}}{2}\) so \(\theta=arcsin(\frac{e^{x}}{2})\). Hence:

\[f(x)=arcsin(\frac{e^{x}}{2})+C\]

To determine \(C\), we use the given point \((0,\frac{\pi}{2})\), which satisfies \(f(x)\):

\[f(0)=arcsin(\frac{e^{0}}{2})+C=arcsin(\frac{1}{2})+C\]

From trigonometry, \(arcsin(\frac{1}{2})=\frac{\pi}{6}\). Therefore:

\[f(0)=\frac{\pi}{6}+C=\frac{\pi}{2}\]

Solving for \(C\):

\[C=\frac{\pi}{2}-\frac{\pi}{6}=\frac{3\pi}{6}-\frac{\pi}{6}=\frac{\pi}{3}\]

Thus, the constant of integration is:

\[\frac{\pi}{3}\]