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  • explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 32 maths

explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 32 maths

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Answer:\frac{1}{2}\{\log (\operatorname{cosec} x-\cot x)\}^{2}+c

Hint:Use substitution method to solve this integral.

Given:   \int \operatorname{cosec} x \cdot \log (\operatorname{cosec} x-\cot x) d x

Solution:

        \begin{aligned} &\text { Let } I=\int \operatorname{cosec} x \cdot \log (\operatorname{cosec} x-\cot x) d x \\ &\text { Put } \log (\operatorname{cosec} x-\cot x)=t \end{aligned}

        \Rightarrow \frac{1}{(\operatorname{cosec} x-\cot x)}\left(-\operatorname{cosec} x \cdot \cot x-\left(-\operatorname{cosec}^{2} x\right)\right) d x=d t

        \Rightarrow \frac{1}{(\operatorname{cosec} x-\cot x)}\left\{\operatorname{cosec}^{2} x-\operatorname{cosec} x \cdot \cot x\right\} d x=d t

        \Rightarrow \frac{1}{(\operatorname{cosec} x-\cot x)} \operatorname{cosec} x\{\operatorname{cosec} x-\cot x\} d x=d t

        \Rightarrow \operatorname{cosec} x d x=d t \Rightarrow d x=\frac{d t}{\operatorname{cosec} x} \text { then }

        \Rightarrow I=\int \operatorname{cosec} x \cdot t \cdot \frac{d t}{\operatorname{cosec} 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 (\operatorname{cosec} x-\cot x)\}^{2}+c \quad \quad[\because t=\log (\operatorname{cosec} x-\cot x)]

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