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If $\int \cos x \log \left(\tan \frac{x}{2}\right) d x$$=\sin x \log \left(\tan \frac{x}{2}\right)+f(x),$ then $f(x)$ is equal to (assuming $c$ is a arbitrary real constant)
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The correct answer is:
$c-x$
Let $I=\int \cos x \log \left(\tan \frac{x}{2}\right)$
$=\log \left(\tan \frac{x}{2}\right) \cdot \sin x-\int \sin x \cdot \frac{1}{\tan \frac{x}{2}} \cdot \sec ^{2} \frac{x}{2} \cdot \frac{1}{2} d x$
$=\sin x \log \left(\tan \frac{x}{2}\right)-\int \sin x \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}} d x$
$=\sin x \cdot \log \left(\tan \frac{x}{2}\right)-\int \frac{\sin x}{\sin x} d x$
$=\sin x \cdot \log \left(\tan \frac{x}{2}\right)-\int 1 d x$
$=\sin x \log \left(\tan \frac{x}{2}\right)-x+c$
$\therefore f(x)=c-x$
$=\log \left(\tan \frac{x}{2}\right) \cdot \sin x-\int \sin x \cdot \frac{1}{\tan \frac{x}{2}} \cdot \sec ^{2} \frac{x}{2} \cdot \frac{1}{2} d x$
$=\sin x \log \left(\tan \frac{x}{2}\right)-\int \sin x \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}} d x$
$=\sin x \cdot \log \left(\tan \frac{x}{2}\right)-\int \frac{\sin x}{\sin x} d x$
$=\sin x \cdot \log \left(\tan \frac{x}{2}\right)-\int 1 d x$
$=\sin x \log \left(\tan \frac{x}{2}\right)-x+c$
$\therefore f(x)=c-x$
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