Question

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

Answer 1

To find: $\int e^x\left(\frac{1+\sin x}{1+\cos x}\right).dx$

We know that

$\sin x=2\sin \frac{x}{2}\cos \frac{x}{2}\text{and}\cos x+1=2{\cos }^2\frac{x}{2}$

Now

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

$=\int e^x\left(\frac{1+2\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)}{2{\cos }^2\left(\frac{x}{2}\right)}\right).dx$

$=\int e^x\left(\frac{{\sin }^2\left(\frac{x}{2}\right)+{\cos }^2\left(\frac{x}{2}\right)+2\sin \left(\frac{x}{2}\right).\cos \frac{x}{2}}{2{\cos }^2\left(\frac{x}{2}\right)}\right).dx$
$[\because {\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}=1]$

$=\int e^x\left[\frac{{(\sin \frac{x}{2}+\cos \frac{x}{2})}^2}{2{\cos }^2\left(\frac{x}{2}\right)}\right].dx$

$=\frac{1}{2}\int e^x{\left(\frac{\sin \frac{x}{2}+\cos \frac{x}{2}}{\cos \frac{x}{2}}\right)}^2.dx$

$=\frac{1}{2}\int e^x{(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}+\frac{\cos \frac{x}{2}}{\cos \frac{x}{2}})}^2.dx$

$=\frac{1}{2}\int e^x{(\tan \frac{x}{2}+1)}^2.dx$

$=\frac{1}{2}\int e^x({\tan }^2\frac{x}{2}+1+2\tan \frac{x}{2}).dx$

$=\frac{1}{2}\int e^x({\sec }^2\frac{x}{2}+2\tan \frac{x}{2}).dx$

Now, let

$2\tan \frac{x}{2}=f\left(x\right)\Rightarrow f^{\prime }\left(x\right)=2\left({\sec }^2\frac{x}{2}\right).\frac{1}{2}$

$={\sec }^2\frac{x}{2}$

Hence,

$\frac{1}{2}\int e^x({\sec }^2\frac{x}{2}+2\tan \frac{x}{2}).dx$

$=\frac{1}{2}\int e^x\left(f^{\prime }\right(x)+f(x\left)\right).dx$

$=\frac{1}{2}{e}^{x}f\left(x\right)+C$

$=\frac{1}{2}{e}^x\left(2\tan \frac{x}{2}\right)+C$

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

Where $C$ is constant of integration.