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