The value of the integral 48 π ^ 4 ∫ _ 0 ^ π ( 3π x ^ 2 2 - x ^ 3 )...

Let $I = \dfrac{48}{{\pi}^{4}}{∫}_{0}^{\pi}\left(\dfrac{3\pi}{2}{x}^{2} - {x}^{3}\right)\dfrac{\sin x}{1 + \cos^{2}x}dx \ldots \left(1\right)$

Using ${∫}_{a}^{b}f\left(x\right)dx = {∫}_{a}^{b}f ( a + b - x ) dx$

We get $I = \dfrac{48}{{\pi}^{4}}{∫}_{0}^{\pi}\left(\dfrac{3\pi}{2}{\left(\pi - x\right)}^{2} - {\left(\pi - x\right)}^{3}\right)\dfrac{\sin \left(0 + \pi - x\right)}{1 + \cos^{2}\left(0 + \pi - x\right)}dx$

$I = \dfrac{48}{{\pi}^{4}}{∫}_{0}^{\pi}\left(\dfrac{3{\pi}^{3}}{2} - 3{\pi}^{2}x + \dfrac{3\pi{x}^{2}}{2} - {\pi}^{3} + {x}^{3} + 3{\pi}^{2}x - 3\pi{x}^{2}\right)\dfrac{\sin x}{1 + \cos^{2}x}dx$ ..............equation $2$

Now adding equation (1) + equation (2)

$2I = \dfrac{48}{{\pi}^{4}}{∫}_{0}^{\pi}\left(\dfrac{3{\pi}^{3}}{2} - {\pi}^{3}\right)\dfrac{\sin x}{1 + \cos^{2}x}dx$

$2I = \dfrac{48}{{\pi}^{4}} \times \dfrac{{\pi}^{3}}{2}{∫}_{0}^{\pi}\dfrac{\sin x}{1 + \cos^{2}x}dx$

$2I = \dfrac{24}{\pi} \times 2 \times {∫}_{0}^{\dfrac{\pi}{2}}\dfrac{\sin x}{1 + \cos^{2}x}dx$

Let $\cos x = t ⇒ - sinxdx = dt$

So $I = \dfrac{24}{\pi} \times {∫}_{1}^{0}\dfrac{- dt}{1 + {t}^{2}}$

$⇒ I = \dfrac{- 24}{\pi}{\left[\tan^{1}t\right]}_{1}^{0}$

$⇒ I = - \dfrac{24}{\pi}\left[0 - \dfrac{\pi}{4}\right] = 6$