The integral to be evaluated is:
\[f(x)=\int\frac{e^{x}}{\sqrt{4-e^{2x}}}dx\]
The denominator is simplified as:
\[\sqrt{4-e^{2x}}=\sqrt{(2)^{2}-(e^{x})^{2}}\]
This form suggests the substitution \(e^{x}=2~sin~\theta\). Consequently, \(e^{2x}=4~sin^{2}\theta\). The differential \(dx\) becomes:
\[dx=\frac{2~cos~\theta}{e^{x}}d\theta=\frac{2~cos~\theta}{2~sin~\theta}d\theta=cot~\theta~d\theta\]
Substituting these into the integral yields:
\[f(x)=\int\frac{2~sin~\theta}{\sqrt{4-4~sin^{2}\theta}}cot~\theta~d\theta\]
Using the identity \(\sqrt{4-4~sin^{2}\theta}=2~cos~\theta\):
\[f(x)=\int\frac{2~sin~\theta}{2~cos~\theta}cot~\theta~d\theta=\int d\theta\]
The integral simplifies to:
\[f(x)=\theta+C\]
Reverting to the original variable using \(e^{x}=2~sin~\theta\), we get \(sin~\theta=\frac{e^{x}}{2}\), which implies \(\theta=arcsin(\frac{e^{x}}{2})\). Therefore, the function is:
\[f(x)=arcsin(\frac{e^{x}}{2})+C\]
The constant of integration \(C\) is determined using the given point \((0,\frac{\pi}{2})\) that satisfies \(f(x)\):
\[f(0)=arcsin(\frac{e^{0}}{2})+C=arcsin(\frac{1}{2})+C\]
From trigonometric values, \(arcsin(\frac{1}{2})=\frac{\pi}{6}\). Substituting this, we get:
\[f(0)=\frac{\pi}{6}+C=\frac{\pi}{2}\]
Solving for \(C\):
\[C=\frac{\pi}{2}-\frac{\pi}{6}=\frac{3\pi}{6}-\frac{\pi}{6}=\frac{\pi}{3}\]
The constant of integration is found to be:
\[\frac{\pi}{3}\]