int_{-1}^{1} sin^3 x cos^2 x , dx = | vedclass

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(A) Let $f(x) = \sin^3 x \cos^2 x$. Check if the function is even or odd by evaluating $f(-x)$: $f(-x) = \sin^3(-x) \cos^2(-x) = (-\sin x)^3 (\cos x)^

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(A) Let $f(x) = \sin^3 x \cos^2 x$. Check if the function is even or odd by evaluating $f(-x)$: $f(-x) = \sin^3(-x) \cos^2(-x) = (-\sin x)^3 (\cos x)^2 = -\sin^3 x \cos^2 x = -f(x)$. Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function. According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) \, dx = 0$. Therefore,$\int_{-1}^{1} \sin^3 x \cos^2 x \, dx = 0$.

$\int_{-1}^{1} \sin^3 x \cos^2 x \, dx = $

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