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Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 23

Answers (1)

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Answer:

0

Given:

\int_{-\pi}^{\pi} \sin ^{3} x \cos ^{2} x \; d x

Hint:

This equation is solve by \int_{-a}^{a} f(x) d x
 

Explanation:  

\begin{aligned} &\int_{-a}^{a} f(x) d x \\\\ &f(-x)=\sin ^{3}(-x) \cdot \cos ^{2}(-x) \end{aligned}

\begin{aligned} &=-(\sin x)^{3} \cdot \cos ^{2} x \\\\ &=-\sin ^{3} x \cdot \cos ^{2} x \\\\ &=-f(x) \end{aligned}

f(x) is odd function.

\int_{-\pi}^{\pi} \sin ^{3} x \cos ^{2} x \; d x=0

 

 

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