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$\int\left[\log (1+\cos x)-x \tan \left(\frac{x}{2}\right)\right] d x=$
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Verified Answer
The correct answer is:
$x \log |1+\cos x|+c$
$$
\begin{aligned}
I &=\int\left[\log (1+\cos x)-x \tan \left(\frac{x}{2}\right)\right] d x \\
I &=\int \log (1+\cos x) \cdot 1 d x-\int x \tan \frac{x}{2} d x \\
&=x \log (1+\cos x)-\int \frac{(-\sin x)(x)}{1+\cos x} d x-\int x \tan \frac{x}{2}
\end{aligned}
$$
$=x \log (1+\cos x)+\int \frac{x\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)}{2 \cos ^{2} \frac{x}{2}} d x-\int x \tan \frac{x}{2} d x$
$=x \log (1+\cos x)+\int x \tan \frac{x}{2} d x-\int x \tan \frac{x}{2} d x$
$=x \log (1+\cos x)+c$
\begin{aligned}
I &=\int\left[\log (1+\cos x)-x \tan \left(\frac{x}{2}\right)\right] d x \\
I &=\int \log (1+\cos x) \cdot 1 d x-\int x \tan \frac{x}{2} d x \\
&=x \log (1+\cos x)-\int \frac{(-\sin x)(x)}{1+\cos x} d x-\int x \tan \frac{x}{2}
\end{aligned}
$$
$=x \log (1+\cos x)+\int \frac{x\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)}{2 \cos ^{2} \frac{x}{2}} d x-\int x \tan \frac{x}{2} d x$
$=x \log (1+\cos x)+\int x \tan \frac{x}{2} d x-\int x \tan \frac{x}{2} d x$
$=x \log (1+\cos x)+c$
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