int_{-pi / 4}^{pi / 4} x^3 sin ^4(x) d x= | vedclass

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(A) Let $I = \int_{-\pi / 4}^{\pi / 4} x^3 \sin^4(x) dx$. We know that for a definite integral,$\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is an odd functio

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$0$

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(A) Let $I = \int_{-\pi / 4}^{\pi / 4} x^3 \sin^4(x) dx$. We know that for a definite integral,$\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is an odd function,i.e.,$f(-x) = -f(x)$. Here,$f(x) = x^3 \sin^4(x)$. Calculating $f(-x)$: $f(-x) = (-x)^3 \sin^4(-x) = -x^3 (\sin(x))^4 = -x^3 \sin^4(x) = -f(x)$. Since $f(x)$ is an odd function,the integral over the symmetric interval $[-\pi/4, \pi/4]$ is $0$. Therefore,$I = 0$.

$\int_{-\pi / 4}^{\pi / 4} x^3 \sin ^4(x) d x=$

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