The value of the integral frac{48}{pi^{4}} in | vedclass

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(A) Let $I = \frac{48}{\pi^{4}} \int_{0}^{\pi} x^{2} \left(\frac{3 \pi}{2} - x\right) \frac{\sin x}{1 + \cos^{2} x} dx$.Using the property $\int_{0}^{

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$6$

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(A) Let $I = \frac{48}{\pi^{4}} \int_{0}^{\pi} x^{2} \left(\frac{3 \pi}{2} - xight) \frac{\sin x}{1 + \cos^{2} x} dx$.Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we have:$I = \frac{48}{\pi^{4}} \int_{0}^{\pi} (\pi-x)^{2} \left(\frac{3 \pi}{2} - (\pi-x)ight) \frac{\sin x}{1 + \cos^{2} x} dx$$I = \frac{48}{\pi^{4}} \int_{0}^{\pi} (\pi-x)^{2} \left(\frac{\pi}{2} + xight) \frac{\sin x}{1 + \cos^{2} x} dx$.Adding the two expressions for $I$,we get:$2I = \frac{48}{\pi^{4}} \int_{0}^{\pi} \frac{\sin x}{1 + \cos^{2} x} [x^{2}(\frac{3\pi}{2}-x) + (\pi-x)^{2}(\frac{\pi}{2}+x)] dx$.Simplifying the term inside the bracket:$x^{2}(\frac{3\pi}{2}-x) + (\pi^{2}-2\pi x+x^{2})(\frac{\pi}{2}+x) = \frac{3\pi x^{2}}{2} - x^{3} + \frac{\pi^{3}}{2} + \pi^{2}x - \pi^{2}x - 2\pi x^{2} + \frac{x^{2}\pi}{2} + x^{3} = \frac{\pi^{3}}{2}$.Thus,$2I = \frac{48}{\pi^{4}} \int_{0}^{\pi} \frac{\sin x}{1 + \cos^{2} x} \cdot \frac{\pi^{3}}{2} dx = \frac{24}{\pi} \int_{0}^{\pi} \frac{\sin x}{1 + \cos^{2} x} dx$.Let $\cos x = t$,then $-\sin x dx = dt$. When $x=0, t=1$; when $x=\pi, t=-1$.$2I = \frac{24}{\pi} \int_{1}^{-1} \frac{-dt}{1+t^{2}} = \frac{24}{\pi} \int_{-1}^{1} \frac{dt}{1+t^{2}} = \frac{24}{\pi} [	an^{-1} t]_{-1}^{1} = \frac{24}{\pi} (\frac{\pi}{4} - (-\frac{\pi}{4})) = \frac{24}{\pi} \cdot \frac{\pi}{2} = 12$.Therefore,$I = 6$.

The value of the integral $\frac{48}{\pi^{4}} \int_{0}^{\pi} \left(\frac{3 \pi x^{2}}{2} - x^{3}\right) \frac{\sin x}{1 + \cos^{2} x} dx$ is equal to

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