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Answer:

$x \tan \frac{x}{2}+C$

Given:

$\int \frac{x+\sin x}{1+\cos x} d x$

Hint:

Using integration by parts and $\int \sec ^{2} \theta d \theta$ .

Explanation:

Let $\mathrm{I}=\int \frac{x+\sin x}{1+\cos x} d x$

\begin{aligned} &=\int \frac{x}{1+\cos x} d x+\int \frac{\sin x}{1+\cos x} d x \\ &=\int \frac{x}{2 \cos ^{2} \frac{x}{2}} d x+\int \frac{2 \sin \frac{x}{2} \cos ^{x}}{2 \cos ^{2} \frac{x}{2}} d x \end{aligned}                                                        $\left[\because 1+\cos 2 \theta=2 \cos ^{2} \theta ; \sin 2 \theta=2 \sin \theta \cos \theta\right]$

$=\frac{1}{2} \int x \sec ^{2} \frac{x}{2} d x+\int \tan \frac{x}{2} d x$                                                                $\left[\text { Put } \frac{x}{2}=t \Rightarrow x=2 t \Rightarrow d x=2 d t\right]$

\begin{aligned} &=\frac{1}{2} \int 2 t \sec ^{2} t .2 d t+\int \tan t \cdot 2 d t \\ &=2 \int t \sec ^{2} t d t+2 \int \tan t d t \\ &=2\left[t . t a n t-\int \tan t d t\right]+2 \int \tan t d t \ldots\left\{\int u . v d x=u \int v d x-\int \frac{d u}{d x} \int v d x d x\right\} \\ &=2 t . \tan t-2 \int \tan t d t+2 \int \tan t d t \\ &=2 t \tan t+C \\ &=2\left(\frac{x}{2}\right) \tan \frac{x}{2}+C \\ &=x \tan \frac{x}{2}+C \end{aligned}

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