Question

The value of $\int e^x\left[\frac{1+\sin x}{1+\cos x}\right]dx$ is

• A. $\frac{1}{2}{e}^x\sec \frac{x}{2}+C$
• B. $e^x\sec \frac{x}{2}+C$
• C. $\frac{1}{2}{e}^x\tan \frac{x}{2}+C$
• D. $e^x\tan \frac{x}{2}+C$

Answer 1

The correct option is D $e^x\tan \frac{x}{2}+C$

Let

$I=\int e^x\left[\frac{1+\sin x}{1+\cos x}\right]dx$

$=\int \{\frac{e^x}{(1+\cos x)}+\frac{e^x\sin x}{(1+\cos x)}\}dx$

$=\int \frac{e^x}{2{\cos }^2\frac{x}{2}}dx+\int \frac{e^x\cdot 2\sin \frac{x}{2}\cdot \cos \frac{x}{2}}{2{\cos }^2\frac{x}{2}}\cdot dx$

$=\frac{1}{2}\int e^x\cdot {\sec }^2\frac{x}{2}\cdot dx+\int \underset{II}{e^x}\underset{I}{\tan }\frac{x}{2}\cdot dx$

$=\frac{1}{2}\int e^x\cdot {\sec }^2\frac{x}{2}\cdot dx+\{\tan \frac{x}{2}\cdot e^x-\int \frac{1}{2}\cdot {\sec }^2\frac{x}{2}\cdot e^{x}dx\}$

(using integral by parts)
$=\frac{1}{2}\int e^x\cdot {\sec }^2\frac{x}{2}dx+e^x\cdot \tan \frac{x}{2}-\frac{1}{2}\int e^x\cdot {\sec }^2\frac{x}{2}\cdot dx$

$=e^x\cdot \tan \frac{x}{2}+C$