The value of the integral ∫ _ 0 ^ (π)/(4) xdx ^ 4 (2x) + ^ 4 (2x)...

Let, $\mathop{I = ∫}_{0}^{\dfrac{\pi}{4}}\dfrac{xdx}{\sin^{4}\left(2x\right) + \cos^{4}\left(2x\right)}$

Let, $2x = t$ then $dx = \dfrac{1}{2}dt$

$⇒ I = \dfrac{1}{4}\mathop{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{tdt}{\sin^{4}t + \cos^{4}t}$

$⇒ I = \dfrac{1}{4}\mathop{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{\left(\dfrac{\pi}{2} - t\right)dt}{\sin^{4}\left(\dfrac{\pi}{2} - t\right) + \cos^{4}\left(\dfrac{\pi}{2} - t\right)}$

$⇒ I = \dfrac{1}{4}\mathop{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{\dfrac{\pi}{2}dt}{\sin^{4}t + \cos^{4}t} - I$

$⇒ 2I = \dfrac{\pi}{8}\mathop{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{dt}{\sin^{4}t + \cos^{4}t}$

$⇒ 2I = \dfrac{\pi}{8}\mathop{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{\sec^{4}tdt}{\tan^{4}t + 1}$

Let, $\tan t = y$ then $\sec^{2}tdt = dy$

$⇒ 2I = \dfrac{\pi}{8}\mathop{∫}_{0}^{\infty}\dfrac{\left(1 + {y}^{2}\right)dy}{1 + {y}^{4}}$

$⇒ I = \dfrac{\pi}{16}\mathop{∫}_{0}^{\infty}\dfrac{1 + \dfrac{1}{{y}^{2}}}{{y}^{2} + \dfrac{1}{{y}^{2}}}dy$

Putting, $y - \dfrac{1}{y} = p$

$⇒ I = \dfrac{\pi}{16}\mathop{∫}_{- \infty}^{\infty}\dfrac{dp}{{p}^{2} + {\left(\sqrt{2}\right)}^{2}}$

$⇒ I = \dfrac{\pi}{16\sqrt{2}}{\left[\tan^{- 1}\left(\dfrac{p}{\sqrt{2}}\right)\right]}_{- \infty}^{\infty}$

$⇒ I = \dfrac{{\pi}^{2}}{16\sqrt{2}}$

$⇒ I = \dfrac{\sqrt{2}{\pi}^{2}}{32}$