Q. $\int_{}^{}{\frac{e^{x}(1 + \sin x)}{1 + \cos x}dx}$ is equal to
A) $e^{x}\tan{\left( \frac{x}{2} \right) + c}$
B) $e^{x}\tan{x + c}$
C) $e^{x}\left( \frac{1 + \sin x}{1 - \cos x} \right) + c$
D) $c - e^{x}\cot\left( \frac{x}{2} \right)$
Explanation
(a)
Let $I = \int_{}^{}{\frac{e^{x}(1 + \sin x)}{1 + \cos x}dx}$
$= \int_{}^{}\frac{1}{2}e^{x}\sec^{2}\frac{x}{2}dx + \int_{}^{}e^{x}\tan\frac{x}{2}dx$
$= \frac{1}{2}\left\lbrack 2e^{x}\tan{\frac{x}{2} - \int_{}^{}{2e^{x}}}\tan\frac{x}{2}dx \right\rbrack + \int_{}^{}e^{x}\tan\frac{x}{2}dx$
$= e^{x}\tan\frac{x}{2} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.