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Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 2

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Hint:You must know about the odd and even function.

Given: \int_{-\pi}^{\pi} \sin ^{3} x \cos ^{2} x\; dx

Solution:  

Here, firstly we have to check the function is odd or even:

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

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

f (x) is odd function.

As for odd function \int_{-a}^{a} f(x)=0

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

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