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provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise  18.9 question 18

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Answer:\frac{1}{6} \sin ^{6} x+c

Hint: Use substitution method to solve this integral.

Given: \int \sin ^{5} x \cos x\; d x


        \begin{aligned} &\text { Let } I=\int \sin ^{5} x \cos x d x \\ &\text { Put } \sin x=t \Rightarrow \cos x d x=d t \text { then } \end{aligned}

        \begin{aligned} I &=\int t^{5} d t \\ &=\frac{t^{5+1}}{5+1}+\mathrm{c} \quad \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \end{aligned}

           \begin{aligned} &=\frac{t^{6}}{6}+c \\ &=\frac{1}{6} \sin ^{6} x+c \quad \quad[\because t=\sin x] \end{aligned}

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