Evaluate the following integrals:

Solution:

Let I $=\int e^{x}\left(\frac{1+\sin x}{1+\cos x}\right) d x$

We know that, $\sin ^{2} x+\cos ^{2} x=1$ and $\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2}$

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

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

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

$=\frac{1}{2} e^{x}\left[\tan \frac{x}{2}+1\right]^{2}$

$=\frac{1}{2} e^{x}\left[1+\tan \frac{x}{2}\right]^{2}$

$=\frac{1}{2} e^{x}\left[1+\tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}\right]$

$=\frac{1}{2} e^{x}\left[\sec ^{2} \frac{x}{2}+2 \tan \frac{x}{2}\right]$

$=e^{x}\left[\frac{1}{2} \sec ^{2} \frac{x}{2}+\tan \frac{x}{2}\right] \ldots \ldots(1)$

Let $\tan \frac{x}{2}=f(x)$

$f^{\prime}(x)=\frac{1}{2} \sec ^{2} \frac{x}{2}$

We know that,

$\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+c$

From equation(1), we obtain

$\int e^{x}\left(\frac{1+\sin x}{1+\cos x}\right) d x=e^{x} \tan \frac{x}{2}+c$