Q62 / 150 · MHT CET 2012 · Mathematics
The value of $\int e^{x}\left[\frac{1+\sin x}{1+\cos x}\right] d x$ is
- A$\frac{1}{2} e^{x} \sec \frac{x}{2}+C$
- B$e^{x} \sec \frac{x}{2}+C$
- C$\frac{1}{2} e^{x} \tan \frac{x}{2}+C$
- Correct answerD$e^{x} \tan \frac{x}{2}+C$
Solution
Let $I=\int e^{x}\left[\frac{1+\sin x}{1+\cos x}\right] d x$
$\begin{aligned}=& \int\left\{\frac{e^{x}}{(1+\cos x)}+\frac{e^{x} \sin x}{(1+\cos x)}\right\} d x \\=& \int \frac{e^{x}}{2 \cos ^{2} \frac{x}{2}} d x+\int \frac{e^{x} \cdot 2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}} \cdot d x \\=& \frac{1}{2} \int e^{x} \cdot \sec ^{2} \frac{x}{2} \cdot d x+\int \underset{\mathbb{I}}{e^{x} \tan } \frac{x}{2} \cdot d x \\=& \frac{1}{2} \int e^{x} \cdot \sec ^{2} \frac{x}{2} \cdot d x \\+\left\{\tan \frac{x}{2} \cdot e^{x}-\int \frac{1}{2} \cdot \sec ^{2} \frac{x}{2} \cdot e^{x} d x\right\} \end{aligned}$
(using integral by parts)
$=\frac{1}{2} \int e^{x} \cdot \sec ^{2} \frac{x}{2} d x+e^{x} \cdot \tan \frac{x}{2}-\frac{1}{2}$
$\int e^{x} \cdot \sec ^{2} \frac{x}{2} \cdot d x$
$=e^{x} \cdot \tan \frac{x}{2}+C$
Hence, the correct answer is option: 'D'.