|120 ^3 _0^ x^2 x x ^4x+ ^4xdx| ________.

Question

$|\frac{120}{{\pi }^{3}}\int _{0}^{\pi }\frac{{x}^{2}\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx|$ \_\_\_\_\_\_\_\_.

Answer: 15

Accepted Answer

$\begin{gathered} \int _{0}^{\pi }\frac{{x}^{2}\sin x\cdot \cos x}{{\sin }^{4}x+{\cos }^{4}x}dx \ =\int _{0}^{\frac{\pi }{2}}\frac{\sin x\cdot \cos x}{{\sin }^{4}x+{\cos }^{4}x}({x}^{2}-(\pi -x{)}^{2})dx \ =\int _{0}^{\frac{\pi }{2}}\frac{\sin x\cdot \cos x(2\pi x-{\pi }^{2})}{{\sin }^{4}x+{\cos }^{4}x} \ =2\pi \int _{0}^{\frac{\pi }{2}}\frac{x\sin x\cos x}{{\sin }^{4}x+\cos 4x}dx-{\pi }^{2}\int _{0}^{\frac{\pi }{2}}\frac{\sin x\cos x}{{\sin }^{4}x+\cos 4x}dx \ =2\pi \cdot \frac{\pi }{4}\int _{0}^{\frac{\pi }{2}}\frac{\sin x{\cos }^{4}x}{{\sin }^{4}x+{\cos }^{4}x}dx-{\pi }^{2}\int _{0}^{\frac{\pi }{2}}\frac{\sin x{\cos }^{4}x}{{\sin }^{4}x+{\cos }^{4}x}dx \end{gathered}$

$\begin{gathered} =-\frac{{\pi }^{2}}{2}\int _{0}^{\frac{\pi }{2}}\frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx \ =-\frac{{\pi }^{2}}{2}\int _{0}^{\frac{\pi }{2}}\frac{\sin x\cos xdx}{1-2{\sin }^{2}x	imes {\cos }^{2}x} \ =-\frac{{\pi }^{2}}{2}\int _{0}^{\frac{\pi }{2}}\frac{\sin 2x}{2-{\sin }^{2}2x}dx \ =-\frac{{\pi }^{2}}{2}\int _{0}^{\frac{\pi }{2}}\frac{\sin 2x}{1+{\cos }^{2}2x}dx \end{gathered}$

Let $\cos 2x=t$

Solutions · 1