int_{-pi}^{pi} (1-x^2) sin x cdot cos^2 x , d | vedclass

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(D) Let $f(x) = (1-x^2) \sin x \cdot \cos^2 x$. We check if the function is even or odd by evaluating $f(-x)$: $f(-x) = (1-(-x)^2) \sin(-x) \cdot \cos

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(D) Let $f(x) = (1-x^2) \sin x \cdot \cos^2 x$. We check if the function is even or odd by evaluating $f(-x)$: $f(-x) = (1-(-x)^2) \sin(-x) \cdot \cos^2(-x)$ Since $(-x)^2 = x^2$,$\sin(-x) = -\sin x$,and $\cos(-x) = \cos x$: $f(-x) = (1-x^2) (-\sin x) \cdot (\cos x)^2$ $f(-x) = -(1-x^2) \sin x \cdot \cos^2 x = -f(x)$. 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,$\int_{-\pi}^{\pi} (1-x^2) \sin x \cdot \cos^2 x \, dx = 0$.

$\int_{-\pi}^{\pi} (1-x^2) \sin x \cdot \cos^2 x \, dx$ is equal to:

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