Question

$\int e^x\left[\frac{1+\sin x}{1+\cos x}\right]dx$ is equal to
(where $c$ is integration constant)

• A. $e^x\tan \frac{x}{2}+c$
• B. $e^x\frac{\tan x}{2}+c$
• C. $\frac{e^x}{1+\cos x}+c$
• D. $e^x[1+\sin \left(\frac{x}{2}\right)]+c$

Answer 1

The correct option is A $e^x\tan \frac{x}{2}+c$

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

We know
$1+\cos 2x=2{\cos }^{2}x\sin 2x=2\sin x\cos x$
So,
$I=\int e^x[\frac{1}{2}{\sec }^2\left(\frac{x}{2}\right)+\tan \left(\frac{x}{2}\right)]dx\underset{f^{\prime }\left(x\right)}{\downarrow }\underset{f\left(x\right)}{\downarrow }\therefore I=e^x\tan \left(\frac{x}{2}\right)+c[\because \int e^x\left[f\right(x)+f^{\prime }(x\left)\right]dx=e^{x}f\left(x\right)+c]$