The value of the integral _0^ /4xdx ^4(2x)+ ^4(2x) equals :

Question

The value of the integral $\int _{0}^{\pi /4}\frac{xdx}{{\sin }^{4}(2x)+{\cos }^{4}(2x)}$ equals :

Accepted Answer

Take $I=\int _{0}^{\pi /4}\frac{xdx}{{\sin }^{4}(2x)+{\cos }^{4}(2x)}$

Let $2x=t$

$2dx=dt$

$dx=\frac{dt}{2}$

$\begin{gathered} I=\int _{0}^{\pi /2}\frac{t/2\cdot 1/2dt}{{\sin }^{4}t+{\cos }^{4}t} \ I=\frac{1}{4}\int _{0}^{\pi /2}\frac{tdt}{{\sin }^{4}t+{\cos }^{4}t} \ =\frac{1}{4}\int _{0}^{\pi /2}\frac{(\frac{\pi }{2}-t)dt}{{\sin }^{4}(\pi /2-t)+{\cos }^{4}(\pi /2-t)} \ =\frac{1}{4}\int _{0}^{\pi /2}\frac{(\frac{\pi }{2}-t)}{{\sin }^{4}t+{\cos }^{4}t} \end{gathered}$

$\begin{gathered} 2I=\frac{1}{4}\int _{0}^{\pi /2}\frac{\frac{\pi }{2}}{{\sin }^{4}t+{\cos }^{4}t}dt \ 2I=\frac{\pi }{8}\int _{0}^{\pi /2}\frac{dt}{{\sin }^{4}t+{\cos }^{4}t} \ 2I=\frac{\pi }{8}\int _{0}^{\pi /2}\frac{{\sec }^{4}t}{1+{	an }^{4}t}dt \end{gathered}$

$\begin{gathered} =y \ {\sec }^{2}tdt=dy \ 2I=\frac{\pi }{8}\int _{0}^{\infty}\frac{(1+{y}^{2})dy}{1+{y}^{4}} \ =\frac{\pi }{8}\int _{0}^{\infty}\frac{1+\frac{1}{{y}^{2}}}{{y}^{2}+\frac{1}{{y}^{2}}-2+2}dy \ =\frac{\pi }{8}\int _{0}^{\infty}\frac{(1+\frac{1}{{y}^{2}})dy}{2+{(y-\frac{1}{y})}^{2}} \end{gathered}$

$\begin{gathered} y-\frac{1}{y}=4 \ 2I=\frac{\pi }{8}\int _{-\infty}^{\infty}\frac{du}{2+{u}^{2}} \ =\frac{\pi }{8\sqrt{2}}{[{	an }^{-1}\frac{y}{\sqrt{2}}]}_{\infty}^{-\infty} \ =\frac{\sqrt{2}{\pi }^{2}}{32} \end{gathered}$

Solutions · 1