$\cot x \log \sin x$

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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