Question

Integrate the function $I={\int }_0^{2\pi }\frac{xsi{n}^{2n}x}{si{n}^{2x}x+co{s}^{2n}x}$

Answer 1

$I={\int }_0^{2\pi }\frac{xsi{n}^{2n}x}{si{n}^{2x}x+co{s}^{2n}x}I={\int }_0^{2\pi }\frac{(2\pi +0-x)si{n}^{2n}(2\pi -x)}{si{n}^{2n}(2\pi +x)+co{s}^{2n}(2\pi -x)}I={\int }_0^{2\pi }\frac{2\pi si{n}^{2x}x}{si{n}^{2n}x+co{s}^{2n}x}-{\int }_O^{2\pi }\frac{xsi{n}^{2n}x}{si{n}^{2n}x+co{s}^{2n}x}2I{\int }_0^{2\pi }\frac{2\pi si{n}^{2n}x}{si{n}^{2n}x+co{s}^{2n}x}I=\pi {\int }_0^{2\pi }\frac{si{n}^{2n}x}{si{n}^{2n}x+co{s}^{2n}x}I=4\pi {\int }_0^{2\pi }\frac{si{n}^{2n}}{si{n}^{2n}x+co{s}^{2n}x}....\left(i\right)I=4\pi {\int }_0^{\frac{\pi }{2}}=\frac{si{n}^{2n}(\frac{\pi }{2}-x)}{si{n}^{2n}(\frac{\pi }{2}-x)+co{s}^{2n}(\frac{\pi }{2}-x)}I=4\pi {\int }_0^{\frac{\pi }{2}}\frac{co{s}^{2n}\left(x\right)}{co{s}^{2n}x+si{n}^{2n}x}...\left(ii\right)\left(i\right)+\left(ii\right)2I=4\pi {\int }_0^{\frac{\pi }{2}}\frac{si{n}^{2n}x+co{s}^{2n}x}{si{n}^{2n}x+co{s}^{2n}x}2I=4\pi {\int }_0^{\frac{\pi }{2}}|dx$
$I=2\pi x{|}_0^{\frac{\pi }{2}}I=2\pi (\frac{\pi }{2}-0)I=\frac{2{\pi }^2}{2}I={\pi }^2$