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need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 7

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Answer: \frac{-1}{4} \cot ^{4} x+c

Hint: Use substitution method to solve this integral.

Given: \int \cot ^{3} x \cdot \operatorname{cosec}^{2} x \; d x


        \begin{aligned} &\text { Let } I=\int \cot ^{3} x \cdot \operatorname{cosec}^{2} x \; d x \\ &\text { Put } \cot \mathrm{x}=t \Rightarrow-\operatorname{cosec}^{2} x\; d x=d t \end{aligned}

        \begin{aligned} &\Rightarrow \operatorname{cosec}^{2} x \; d x=-d t \text { then } \\ &I=\int t^{3}(-d t)=-\int t^{3} d t \end{aligned}

           \begin{aligned} &=-\frac{t^{3+1}}{3+1}+c=-\frac{t^{4}}{4}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &=\frac{-1}{4} \cot ^{4} x+c\; \; \; \; \; [\because t=\cot x] \end{aligned}

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