Evaluate the integral: int_0^pi frac{x}{sin x | vedclass

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(A) Let $I = \int_0^\pi \frac{x}{\sin x}(3 \cos^2 x + 2 \sin x + \sin^3 x - 3) dx$. Since $\cos^2 x = 1 - \sin^2 x$,the expression becomes $3(1 - \sin

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$\frac{\pi(5 \pi-12)}{4}$

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(A) Let $I = \int_0^\pi \frac{x}{\sin x}(3 \cos^2 x + 2 \sin x + \sin^3 x - 3) dx$. Since $\cos^2 x = 1 - \sin^2 x$,the expression becomes $3(1 - \sin^2 x) + 2 \sin x + \sin^3 x - 3 = -3 \sin^2 x + 2 \sin x + \sin^3 x$. Dividing by $\sin x$,we get $-3 \sin x + 2 + \sin^2 x$. Thus,$I = \int_0^\pi x(\sin^2 x - 3 \sin x + 2) dx \dots (1)$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$: $I = \int_0^\pi (\pi - x)(\sin^2 x - 3 \sin x + 2) dx \dots (2)$. Adding $(1)$ and $(2)$: $2I = \int_0^\pi \pi(\sin^2 x - 3 \sin x + 2) dx$. $2I = \pi \int_0^\pi (\frac{1 - \cos 2x}{2} - 3 \sin x + 2) dx = \pi \int_0^\pi (\frac{5}{2} - \frac{1}{2} \cos 2x - 3 \sin x) dx$. $2I = \pi [\frac{5}{2}x - \frac{1}{4} \sin 2x + 3 \cos x]_0^\pi$. $2I = \pi [(\frac{5\pi}{2} - 0 + 3(-1)) - (0 - 0 + 3(1))] = \pi [\frac{5\pi}{2} - 3 - 3] = \pi [\frac{5\pi}{2} - 6] = \frac{\pi(5\pi - 12)}{2}$. Therefore,$I = \frac{\pi(5\pi - 12)}{4}$.

Evaluate the integral: $\int_0^\pi \frac{x}{\sin x}(3 \cos^2 x + 2 \sin x + \sin^3 x - 3) dx$

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