The value of \(\int e^{x}\left[\frac{1+\sin x}{1+\cos x}\right] d x\) is

Let \(I=\int e^{x}\left[\frac{1+\sin x}{1+\cos x}\right] d x\)
\(=\int\left\{\frac{e^{x}}{(1+\cos x)}+\frac{e^{x} \sin x}{(1+\cos x)}\right\} d x \)
\(=\int \frac{e^{x}}{2 \cos ^{2} \frac{x}{2}} d x+\int \frac{e^{x} \cdot 2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}} \cdot d x \)
\(=\frac{1}{2} \int e^{x} \cdot \sec ^{2} \frac{x}{2} \cdot d x+\int \underset{\mathbb{I}}{e^{x} \tan } \frac{x}{2} \cdot d x \)
\(=\frac{1}{2} \int e^{x} \cdot \sec ^{2} \frac{x}{2} \cdot d x \)
\(+\left\{\tan \frac{x}{2} \cdot e^{x}-\int \frac{1}{2} \cdot \sec ^{2} \frac{x}{2} \cdot e^{x} d x\right\}\)
(using integral by parts)
\(=\frac{1}{2} \int e^{x} \cdot \sec ^{2} \frac{x}{2} d x+e^{x} \cdot \tan \frac{x}{2}-\frac{1}{2}\)
\(\int e^{x} \cdot \sec ^{2} \frac{x}{2} \cdot d x\)
\(=e^{x} \cdot \tan \frac{x}{2}+C\)