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Integrate the function
$\frac{e^x(1+\sin x)}{1+\cos x}$
$\frac{e^x(1+\sin x)}{1+\cos x}$
Solution:
1137 Upvotes
Verified Answer
$\begin{aligned}
&\mathrm{I}=\int e^x\left[\frac{1+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}\right] d x \\
&=\int e^x\left[\frac{1}{2} \sec ^2 \frac{x}{2}+\tan \frac{x}{2}\right] d x=\int e^x\left[\tan \frac{x}{2}+\frac{1}{2} \sec ^2 \frac{x}{2}\right] d x
\end{aligned}$
Let $e^x \tan \frac{x}{2}=t \Rightarrow e^x\left(\tan \frac{x}{2}+\frac{1}{2} \sec ^2 \frac{x}{2}\right) d x=d t$
$\therefore I=\int d t=t+C=e^x \tan \frac{x}{2}+C$
&\mathrm{I}=\int e^x\left[\frac{1+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}\right] d x \\
&=\int e^x\left[\frac{1}{2} \sec ^2 \frac{x}{2}+\tan \frac{x}{2}\right] d x=\int e^x\left[\tan \frac{x}{2}+\frac{1}{2} \sec ^2 \frac{x}{2}\right] d x
\end{aligned}$
Let $e^x \tan \frac{x}{2}=t \Rightarrow e^x\left(\tan \frac{x}{2}+\frac{1}{2} \sec ^2 \frac{x}{2}\right) d x=d t$
$\therefore I=\int d t=t+C=e^x \tan \frac{x}{2}+C$
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