Question

If $\int \frac{e^x(1+\sin x)dx}{1+\cos x}=e^{x}f(x)+C$, then $f(x)$ is equal to

• A. $\sin \frac{x}{2}$
• B. $\cos \frac{x}{2}$
• C. $\tan \frac{x}{2}$
• D. $\log \frac{x}{2}$

Answer 1

Answer: (C)

$\int e^x\frac{(1+\sin x)}{(1+\cos x)}dx$
$=\int e^x\left[\frac{1}{2}{\sec }^2\frac{x}{2}+\tan \frac{x}{2}\right]dx$
$=\frac{1}{2}\int e^x{\sec }^2\frac{x}{2}dx+\int e^x\tan \frac{x}{2}dx$
$=e^x\tan \frac{x}{2}+C$
But $I=e^{x}f(x)+C$ (given)
$\therefore f(x)=\tan \frac{x}{2}$