Question:
$\cot x \log \sin x$
Solution:
Let $\log \sin x=t$
$\Rightarrow \frac{1}{\sin x} \cdot \cos x d x=d t$
$\therefore \cot x d x=d t$
$\Rightarrow \int \cot x \log \sin x d x=\int t d t$
$=\frac{t^{2}}{2}+C$
$=\frac{1}{2}(\log \sin x)^{2}+C$
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