| 120 π ^ 3 ∫ _ 0 ^ π x ^ 2 x x ^ 4 x + ^ 4 x dx| is equal to...

Let, $I = {∫}_{0}^{\pi}\dfrac{{x}^{2}\sin x \cos x}{\sin^{4}x + \cos^{4}x} d x$

Now using the property ${∫}_{0}^{2a}f\left(x\right) d x = {∫}_{0}^{a}f\left(x\right) d x + {∫}_{0}^{a}f\left(2a - x\right) d x$ we get,

$⇒ I = {∫}_{0}^{\dfrac{\pi}{2}}\dfrac{\sin x \cos x}{\sin^{4}x + \cos^{4}x}\left({x}^{2} - {\left(\pi - x\right)}^{2}\right) d x$

$⇒ I = {∫}_{0}^{\dfrac{\pi}{2}}\dfrac{\sin x \cos x}{\sin^{4}x + \cos^{4}x}\left(2\pi x - {\pi}^{2}\right) d x$

$⇒ I = 2\pi\mathop{\mathop{{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{x \cdot \sin x \cos x}{\sin^{4}x + \cos^{4}x} d x}_{⏟}}_{{I}_{1}} - {\pi}^{2}{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{\sin x \cos x}{\sin^{4}x + \cos^{4}x} d x$

Now, solving ${I}_{1} = {∫}_{0}^{\dfrac{\pi}{2}}\dfrac{x \cdot \sin x \cos x}{\sin^{4}x + \cos^{4}x}dx . . . . \left(1\right)$

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

Adding both equations we get,

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

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

Now, putting the value in $I$ we get,

$⇒ I = 2\pi \cdot \dfrac{\pi}{4}{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{x \cdot \sin x \cos x}{\sin^{4}x + \cos^{4}x}dx - {\pi}^{2}{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{\sin x \cos x}{\sin^{4}x + \cos^{4}x} d x$

$⇒ I = - \dfrac{{\pi}^{2}}{2}{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{\sin x \cos x}{\sin^{4}x + \cos^{4}x} d x$

$⇒ I = - \dfrac{{\pi}^{2}}{2}{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{\sin x \cos x}{1 - 2\sin^{2}x\cos^{2}x} d x$

$⇒ I = - \dfrac{{\pi}^{2}}{2}{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{\dfrac{1}{2}\sin 2x}{1 - \dfrac{1}{2}\sin^{2}2x} d x$

$⇒ I = - \dfrac{{\pi}^{2}}{2}{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{\sin 2x}{2 - \sin^{2}2x} d x$

$⇒ I = - \dfrac{{\pi}^{2}}{2}{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{\sin 2x}{1 + \cos^{2}2x} d x$

Now, let $\cos 2x = t ⇒ - 2\sin 2xdx = dt$

$⇒ I = - \dfrac{{\pi}^{2}}{2}{∫}_{1}^{- 1}\dfrac{- \dfrac{1}{2}}{1 + {t}^{2}} d t$

$⇒ I = - \dfrac{{\pi}^{2}}{4}{∫}_{- 1}^{1}\dfrac{1}{1 + {t}^{2}} d t$

$⇒ I = - \dfrac{{\pi}^{2}}{4} \cdot \dfrac{\pi}{2} = - \dfrac{{\pi}^{3}}{8}$

Hence, $\dfrac{120}{{\pi}^{3}}\left|{∫}_{0}^{\pi}\dfrac{{x}^{2}\sin x \cos x}{\sin^{4}x + \cos^{4}x} d x\right| = \dfrac{120}{{\pi}^{3}} \times \dfrac{{\pi}^{3}}{8} = 15$