Question

Integrate the function $e^x\left(\frac{1+\sin x}{1+\cos x}\right)$

Answer 1

Given, $e^x\left(\frac{1+\sin x}{1+\cos x}\right)$
$=e^x\left(\frac{{\sin }^2\frac{x}{2}+\cos \frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}}{2{\cos }^2\frac{x}{2}}\right)$
$=\frac{e^x{(\sin \frac{x}{2}+\cos \frac{x}{2})}^2}{2{\cos }^2\frac{x}{2}}$
$=\frac{1}{2}{e}^x\cdot {\left(\frac{\sin \frac{x}{2}+\cos \frac{x}{2}}{\cos \frac{x}{2}}\right)}^2$
$=\frac{1}{2}{e}^x{(1+\tan \frac{x}{2})}^2$
$=\frac{1}{2}{e}^x[1+{\tan }^2\frac{x}{2}+2\tan \frac{x}{2}]$
$=\frac{1}{2}{e}^x[{\sec }^2\frac{x}{2}+2\tan \frac{x}{2}]$
$\frac{e^x(1+\sin x)dx}{(1+\cos x)}=e^x[\frac{1}{2}{\sec }^2\frac{x}{2}+\tan \frac{x}{2}]$ .......... (1)
Let $\tan \frac{x}{2}=f\left(x\right)\Rightarrow f^{\prime }\left(x\right)=\frac{1}{2}{\sec }^2\frac{x}{2}$
It is known that, $\int e^x\left\{f\right(x)+f^{\prime }(x\left)\right\}dx=e^{x}f\left(x\right)+C$
From equation (1), we obtain
$\int \frac{e^x(1+\sin x)}{(1+\cos x)}dx=e^x\tan \frac{x}{2}+C$