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Question Number 200684 by Spillover last updated on 21/Nov/23

$$$$$${\int }_{-4\pi }^{4\pi }\frac{\mid x\mid \sin {}^{2n}x}{\sin {}^{2n}x+\cos {}^{2n}x}dx$$$$$$$$$$

Answered by witcher3 last updated on 22/Nov/23

$$=2{\int }_0^{4\pi }\frac{{\mathrm{xsin}}^{2\mathrm{n}}\left(\mathrm{x}\right)}{{\cos }^{2\mathrm{n}}\left(\mathrm{x}\right)+{\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)}\mathrm{dx}$$$$=2[{\int }_0^{2\pi }\frac{{\mathrm{xsin}}^{2\mathrm{n}}\left(\mathrm{x}\right)}{{\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)+{\cos }^{2\mathrm{n}}\left(\mathrm{x}\right)}+\frac{(2\pi +\mathrm{x}){\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)}{{\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)+{\cos }^{2\mathrm{n}}\left(\mathrm{x}\right)}\mathrm{dx}]$$$$=4{\int }_0^{2\pi }\frac{(\pi +\mathrm{x}){\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)}{{\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)+{\cos }^{2\mathrm{n}}\left(\mathrm{x}\right)}\mathrm{dx}$$$$=4{\int }_0^{\pi }\frac{\pi +\mathrm{x}}{{\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)+{\cos }^{2\mathrm{n}}\left(\mathrm{x}\right)}{\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)+4{\int }_0^{\pi }\frac{\pi +2\pi -\mathrm{x}}{{\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)+{\cos }^{2\mathrm{n}}\left(\mathrm{x}\right)}{\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)$$$$=4{\int }_0^{\pi }\frac{4\pi }{{\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)+{\cos }^{2\mathrm{n}}\left(\mathrm{x}\right)}{\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)$$$$=16\pi [{\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)}{{\cos }^{2\mathrm{n}}\left(\mathrm{x}\right)+{\cos }^{2\mathrm{n}}\left(\mathrm{x}\right)}+{\int }_{\frac{\pi }{2}}^{\pi }\frac{{\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)}{{\cos }^{2\mathrm{n}}\left(\mathrm{x}\right)+{\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)}\mathrm{dx}]$$$$\mathrm{x}\to \frac{\pi }{2}+\mathrm{x}\mathrm{in}2\mathrm{nd}$$$${\cos }^{2\mathrm{n}}(\mathrm{x}+\frac{\pi }{2})={\sin }^{2\mathrm{n}}\left(\mathrm{x}\right);{\sin }^{2\mathrm{n}}(\mathrm{x}+\frac{\pi }{2})={\cos }^{2\mathrm{n}}\left(\mathrm{x}\right)$$$$\mathrm{I}=16\pi {\int }_0^{\frac{\pi }{\mathrm{z}}}\frac{{\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)+{\cos }^{2\mathrm{n}}\left(\mathrm{x}\right)}{{\cos }^{2\mathrm{n}}\left(\mathrm{x}\right)+{\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)}\mathrm{dx}$$$$=16\pi {\int }_0^{\frac{\pi }{2}}\mathrm{dx}=8{\pi }^2$$

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