int_0^{frac{pi}{2}} x^3 sin x , dx = | vedclass

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(D) Let $I = \int_0^{\frac{\pi}{2}} x^3 \sin x \, dx$. Using integration by parts $\int u \, dv = uv - \int v \, du$, let $u = x^3$ and $dv = \sin x \

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$\frac{3 \pi^2}{4} - 6$

Vedclass · Answer

(D) Let $I = \int_0^{\frac{\pi}{2}} x^3 \sin x \, dx$. Using integration by parts $\int u \, dv = uv - \int v \, du$, let $u = x^3$ and $dv = \sin x \, dx$. Then $du = 3x^2 \, dx$ and $v = -\cos x$.$I = [-x^3 \cos x]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} 3x^2 \cos x \, dx = 0 + 3 \int_0^{\frac{\pi}{2}} x^2 \cos x \, dx$.Now, integrate $\int x^2 \cos x \, dx$ by parts again with $u = x^2$ and $dv = \cos x \, dx$. Then $du = 2x \, dx$ and $v = \sin x$.$I = 3 \left( [x^2 \sin x]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} 2x \sin x \, dx \right) = 3 \left( \frac{\pi^2}{4} - 2 \int_0^{\frac{\pi}{2}} x \sin x \, dx \right)$.Finally, integrate $\int x \sin x \, dx$ by parts with $u = x$ and $dv = \sin x \, dx$. Then $du = dx$ and $v = -\cos x$.$\int_0^{\frac{\pi}{2}} x \sin x \, dx = [-x \cos x]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} \cos x \, dx = 0 + [\sin x]_0^{\frac{\pi}{2}} = 1$.Substituting back: $I = 3 \left( \frac{\pi^2}{4} - 2(1) \right) = \frac{3 \pi^2}{4} - 6$.

$\int_0^{\frac{\pi}{2}} x^3 \sin x \, dx =$

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