Mathematics - Definite Integration Question with Solution | TestHub

Let, $\underset{0}{\overset{\frac{\pi }{4}}{I=\int }}\frac{xdx}{{\sin }^4\left(2x\right)+{\cos }^4\left(2x\right)}$

Let, $2x=t$ then $dx=\frac{1}{2}dt$

$\Rightarrow I=\frac{1}{4}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{tdt}{{\sin }^{4}t+{\cos }^{4}t}$

$\Rightarrow I=\frac{1}{4}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\left(\frac{\pi }{2}-t\right)dt}{{\sin }^4\left(\frac{\pi }{2}-t\right)+{\cos }^4\left(\frac{\pi }{2}-t\right)}$

$\Rightarrow I=\frac{1}{4}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\frac{\pi }{2}dt}{{\sin }^{4}t+{\cos }^{4}t}-I$

$\Rightarrow 2I=\frac{\pi }{8}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{dt}{{\sin }^{4}t+{\cos }^{4}t}$

$\Rightarrow 2I=\frac{\pi }{8}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sec }^{4}tdt}{{\tan }^{4}t+1}$

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

$\Rightarrow 2I=\frac{\pi }{8}\underset{0}{\overset{\infty }{\int }}\frac{\left(1+y^2\right)dy}{1+y^4}$

$\Rightarrow I=\frac{\pi }{16}\underset{0}{\overset{\infty }{\int }}\frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy$

Putting, $y-\frac{1}{y}=p$

$\Rightarrow I=\frac{\pi }{16}\underset{-\infty }{\overset{\infty }{\int }}\frac{dp}{p^2+{\left(\sqrt{2}\right)}^2}$

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

$\Rightarrow I=\frac{{\pi }^2}{16\sqrt{2}}$

$\Rightarrow I=\frac{\sqrt{2}{\pi }^2}{32}$