Question

Solve:

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{x\sin 2xdx}{{\cos }^{4}x+{\sin }^{4}x}$

Answer 1

$\begin{array}{c}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{x\sin 2xdx}{{\cos }^{4}x+{\sin }^{4}x}\\ {\left({\sin }^{2}x+{\cos }^{2}x\right)}^2={\sin }^{4}x+{\cos }^{4}x+2{\sin }^{2}x{\cos }^{2}x\\ {\sin }^{4}x+{\cos }^{4}x=1-2{\sin }^{2}x{\cos }^{2}x\\ =1-\left({\sin }^{2}2x\right)/2\\ =\frac{1}{2}+\frac{\left(co{s}^{2}2x\right)}{2}\\ \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{x\sin 2xdx}{\frac{1}{2}+\frac{\left(co{s}^{2}2x\right)}{2}}\\ \cos 2x=t\\ -2\sin 2xdx=dt\\ \underset{1}{\overset{-1}{\int }}\frac{\left(-1\right)dt}{1+t^2}={\left[\left(-1\right){\tan }^{-1}t\right]}_1^{-1}\\ =-{\tan }^{-1}\left(-1\right)+{\tan }^{-1}\left(1\right)\\ =\frac{\pi }{4}+\frac{\pi }{4}=\frac{\pi }{2}\end{array}$