# Show that

Question:

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

Solution:

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

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

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

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

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

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

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

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

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

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

It is known that, $\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+\mathbf{C}$

From equation (1), we obtain

$\int \frac{e^{x}(1+\sin x)}{(1+\cos x)} d x=e^{x} \tan \frac{x}{2}+\mathrm{C}$