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

Answer: (A)

$\frac{\pi}{2}$

Answer 2

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:

$\frac{\pi}{3}$