int_0^pi x sin^3 x cos^2 x , dx = | vedclass

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(A) Let $I = \int_0^\pi x \sin^3 x \cos^2 x \, dx$  ...$(i)$Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$:$I = \int_0^\pi (\pi - x)

Vedclass · Accepted Answer

$\frac{2\pi}{15}$

Vedclass · Answer

(A) Let $I = \int_0^\pi x \sin^3 x \cos^2 x \, dx$  ...$(i)$Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$:$I = \int_0^\pi (\pi - x) \sin^3(\pi - x) \cos^2(\pi - x) \, dx$Since $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$,we have:$I = \int_0^\pi (\pi - x) \sin^3 x \cos^2 x \, dx$  ...(ii)Adding $(i)$ and (ii):$2I = \int_0^\pi \pi \sin^3 x \cos^2 x \, dx = \pi \int_0^\pi (1 - \cos^2 x) \cos^2 x \sin x \, dx$Let $t = \cos x$,then $dt = -\sin x \, dx$. Limits change from $0 	o \pi$ to $1 	o -1$:$2I = \pi \int_1^{-1} (1 - t^2) t^2 (-dt) = \pi \int_{-1}^1 (t^2 - t^4) \, dt$$2I = \pi \left[ \frac{t^3}{3} - \frac{t^5}{5} ight]_{-1}^1 = \pi \left( (\frac{1}{3} - \frac{1}{5}) - (-\frac{1}{3} + \frac{1}{5}) ight) = \pi \left( \frac{2}{15} + \frac{2}{15} ight) = \frac{4\pi}{15}$$I = \frac{2\pi}{15}$

$\int_0^\pi x \sin^3 x \cos^2 x \, dx =$

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