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$\int \frac{x+\sin x}{1+\cos x} d x$ is equal to
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2933 Upvotes
Verified Answer
The correct answer is:
$x \tan \frac{x}{2}+c$
Let $I=\int \frac{x+\sin x}{1+\cos x} d x$
$$
\begin{array}{l}
=\int\left(\frac{x}{2} \sec ^{2} \frac{x}{2}+\tan \frac{x}{2}\right) d x \\
=x \tan \frac{x}{2}-\int \tan \frac{x}{2} d x+\int \tan \frac{x}{2} d x+c
\end{array}
$$
$=x \tan \frac{x}{2}+c$
$$
\begin{array}{l}
=\int\left(\frac{x}{2} \sec ^{2} \frac{x}{2}+\tan \frac{x}{2}\right) d x \\
=x \tan \frac{x}{2}-\int \tan \frac{x}{2} d x+\int \tan \frac{x}{2} d x+c
\end{array}
$$
$=x \tan \frac{x}{2}+c$
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