Question

${\int }_0^{\pi }\frac{xsi{n}^{2n}xdx}{(si{n}^{2n}x+co{s}^{2n}x)}$

Answer 1

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$\begin{array}{c}Byusingpropertyofinterjection\\ I={\int }_0^{\pi }\frac{\left(\pi -x\right){\sin }^{2n}\left(\pi -x\right)dx}{\left[{\sin }^{2n}\left(\pi -x\right)+{\cos }^{2n}\left(\pi -x\right)\right]}\\ \Rightarrow I={\int }_0^{\pi }\frac{\left(\pi -x\right){\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx.......\left(ii\right)\\ Addingequation\left(i\right)and\left(ii\right)\\ \Rightarrow 2I={\int }_0^{\pi }\pi {\left({\sin }^{n}x\right)}^{2}dx\\ 2I={\int }_0^{\pi }\pi {\left({\sin }^{2}x\right)}^{n}dx\\ 2I=\frac{\pi }{2^n}{\int }_0^{\pi }\left(1-n{\cos }^{2}x\right)dx\\ I=\frac{\pi }{2^{(n+1)}}.\left[{\int }_0^{\pi }dx-n{\int }_0^{\pi }\cos 2xdx\right]\\ I=\frac{\pi }{2^{n+1}}\left[\pi -n{\left(\frac{\sin 2x}{2}\right)}_0^{\pi }\right]+c\\ \frac{{\pi }^2}{2^{n+1}}-\frac{n\pi }{2}\left(\frac{n\pi }{2^{n+2}}\right){\left(\sin 2\pi -\sin 0\right)}^{-}+c\\ \frac{{\pi }^2}{2^{x+1}}-\frac{n^{\pi }}{2^{n+1}}\times 0+C\\ \therefore \frac{{\pi }^2}{2^{(n+0)}}\end{array}$