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provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise  18.9 question 30

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Answer: \frac{1}{2}\{\log (\sin x)\}^{2}+c

Hint:Use substitution method to solve this integral.

Given:   \int \cot x \cdot \log (\sin x) d x


        \begin{aligned} &\text { Let } I=\int \cot x \cdot \log (\sin x) d x \\ &\text { Put } \log (\sin x)=t \Rightarrow \frac{1}{\sin x} \cos x\; d x=d t \\ &\Rightarrow \cot x\; d x=d t \Rightarrow d x=\frac{d t}{\cot x} \text { then } \end{aligned}

        \Rightarrow I=\int \cot x \cdot t \cdot \frac{d t}{\cot x}=\int t \; d t

        =\frac{t^{1+1}}{1+1}+c=\frac{t^{2}}{2}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c,\right]

        =\frac{1}{2}\{\log (\sin x)\}^{2}+c \quad[\because t=\log (\sin x)]


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