Evaluate: _0^ x x ^2 x d x - Vidyadip

Question

Evaluate: $\int_0^\pi x \cdot \sin x \cdot \cos ^2 x \cdot d x$

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Answer

$ \text { Let } I =\int_0^\pi x \cdot \sin x \cdot \cos ^2 x \cdot d x \quad \ldots \ldots \text { (i) }$
$\therefore I =\int_0^\pi(\pi-x) \cdot \sin (\pi-x) \cdot[\cos (\pi-x)]^2 d x \quad \ldots . .\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$\therefore I =\int_0^\pi(\pi-x) \cdot \sin x(-\cos x)^2 d x$
$\therefore I =\int_0^\pi(\pi-x) \cdot \sin x \cos ^2 x d x \quad \ldots . \text { (ii) } $
Adding (i) and (ii), we get
$ 2 I =\int_0^\pi x \cdot \sin x \cdot \cos ^2 x d x+\int_0^\pi(\pi-x) \cdot \sin x \cos ^2 x d x$
$=\int_0 \pi(x+\pi-x) \cdot \sin x \cos ^2 x d x$
$\therefore 2 I =\pi \int_0^\pi \sin x \cos ^2 x d x $
Put $\cos x=t$
$\therefore-\sin x d x=d t$
$\therefore \sin xdx =- dt$
When $x=0, t=1$ and when $x=\pi, t=-1$
$ \therefore 2 l =\pi \int_1^{-1} t ^2(- dt )$
$\therefore I =\frac{\pi}{2} \int_{-1}^1 t ^2 dt \ldots \ldots . .\left[\because \int_{ a }^{ b } f (x) d x=-\int_{ b }^{ a } f (x) d x\right]$
$=\frac{\pi}{2} \times 2 \int_0^1 t ^2 dt \ldots . .\left[\because t ^2 \text { is an even function }\right]$
$=\pi\left[\frac{ t ^2}{3}\right]_0^1$
$=\frac{\pi}{3}\left(1^3-0\right)$
$\therefore I =\frac{\pi}{3} $

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