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Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 5 maths textbook solution

Answers (1)

Answer:-\frac{3}{5}(\cos x)^{\frac{5}{3}}+c

Hint: Use substitution method to solve this integral.

Given:\int \sqrt[3]{\cos ^{2} x} \cdot \sin x\; d x

Solution:

         Let I=\int \sqrt[3]{\cos ^{2} x} \cdot \sin x \; d x

        \begin{aligned} &\text { Put } \cos x=t \Rightarrow-\sin x d x=d t \\ &\Rightarrow \sin x d x=-d t \text { then } \\ &I=\int \sqrt[3]{t^{2}}(-d t)=-\int t^{\frac{2}{3}} d t \end{aligned}

        \begin{aligned} &=-\frac{t^{\frac{2}{3}+1}}{\frac{2}{3}+1}+c=-\frac{t^{\frac{5}{3}}}{\frac{5}{3}}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &=-\frac{3}{5}(\cos x)^{\frac{5}{3}}+c \quad[\because t=\cos x] \end{aligned}

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