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Q29  Integrate the functions \cot x \: log \sin x

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Given function  \cot x \: log \sin x,

Assume the \log \sin x =t

 \therefore \frac{1}{\sin x }.\cos x dx =dt

\cot x dx =dt

\implies \int \cot x \log \sin x dx =\int t dt

= \frac{t^2}{2}+C

Now, back substituted the value of t.

= \frac{1}{2}(\log \sin x )^2+C, where C is any constant value.

Posted by

Divya Prakash Singh

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