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0-2pi-xsin-2n-x-sin-2n-x-cos-2n-x-dx-pi-2-prove-

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Question Number 224924 by fantastic last updated on 12/Oct/25
∫_0 ^(2π) ((xsin^(2n) x)/(sin^(2n) x+cos^(2n) x)) dx=π^2 (prove)
$$\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\frac{{x}\mathrm{sin}^{\mathrm{2}{n}} {x}}{\mathrm{sin}^{\mathrm{2}{n}} {x}+\mathrm{cos}^{\mathrm{2}{n}} {x}}\:{dx}=\pi^{\mathrm{2}} \:\left({prove}\right) \\ $$
Commented by fantastic last updated on 12/Oct/25
please give full ans.
$${please}\:{give}\:{full}\:{ans}. \\ $$
Answered by fkwow344 last updated on 14/Oct/25
∫ ((z∙sin^(2n) (2π−z))/(sin^(2n) (2π−z)+cos^(2n) (2π−z)))dz=I ∙∙∙∙(A) Let′s z→2π−z^′ dz=−dz′ z=2π−z′ ∫_0 ^( 2π) (((2π−z′)sin^(2n) (2π−z′))/(sin^(2n) (2π−z′)+cos^(2n) (2π−z′))) dz′= ∫ ((2π∙sin^(2n) (z′))/(sin^(2n) (z′)+cos^(2n) (z′)))−∫ ((z′sin^(2n) (z′))/(sin^(2n) (z′)+cos^(2n) (z′))) ∙∙∙(A) I= ∫ ((2π∙sin^(2n) (z′))/(sin^(2n) (z′)+cos^(2n) (z′)))−I 2I=∫ ((2π∙sin^(2n) (z′))/(sin^(2n) (z′)+cos^(2n) (z′)))dz I=∫ ((π∙sin^(2n) (z′))/(sin^(2n) +cos^(2n) (z′))) dz ∫ ((sin^(2n) (z))/(sin^(2n) (z)+cos^(2n) (z)))=∫ ((cos^(2n) (z))/(sin^(2n) (z)+cos^(2n) (z))) ∫ ((sin^(2n) (z)+cos^(2n) (z))/(sin^(2n) (z)+cos^(2n) (z))) dz=2π. ∫ ((sin^(2n) (z))/(sin^(2n) (z)+cos^(2n) (z)))dz=π ∴π^2
$$\int\:\:\frac{{z}\centerdot\mathrm{sin}^{\mathrm{2}{n}} \left(\mathrm{2}\pi−{z}\right)}{\mathrm{sin}^{\mathrm{2}{n}} \left(\mathrm{2}\pi−{z}\right)+\mathrm{cos}^{\mathrm{2}{n}} \left(\mathrm{2}\pi−{z}\right)}\mathrm{d}{z}={I}\:\centerdot\centerdot\centerdot\centerdot\left({A}\right) \\ $$$$\mathrm{Let}'\mathrm{s}\:{z}\rightarrow\mathrm{2}\pi−{z}^{'} \\ $$$$\mathrm{d}{z}=−\mathrm{d}{z}'\:{z}=\mathrm{2}\pi−{z}' \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \:\:\frac{\left(\mathrm{2}\pi−{z}'\right)\mathrm{sin}^{\mathrm{2}{n}} \left(\mathrm{2}\pi−{z}'\right)}{\mathrm{sin}^{\mathrm{2}{n}} \left(\mathrm{2}\pi−{z}'\right)+\mathrm{cos}^{\mathrm{2}{n}} \left(\mathrm{2}\pi−{z}'\right)}\:\mathrm{d}{z}'= \\ $$$$\int\:\:\frac{\mathrm{2}\pi\centerdot\mathrm{sin}^{\mathrm{2}{n}} \left({z}'\right)}{\mathrm{sin}^{\mathrm{2}{n}} \left({z}'\right)+\mathrm{cos}^{\mathrm{2}{n}} \left({z}'\right)}−\int\:\:\:\frac{{z}'\mathrm{sin}^{\mathrm{2}{n}} \left({z}'\right)}{\mathrm{sin}^{\mathrm{2}{n}} \left({z}'\right)+\mathrm{cos}^{\mathrm{2}{n}} \left({z}'\right)}\:\centerdot\centerdot\centerdot\left({A}\right) \\ $$$${I}=\:\int\:\:\frac{\mathrm{2}\pi\centerdot\mathrm{sin}^{\mathrm{2}{n}} \left({z}'\right)}{\mathrm{sin}^{\mathrm{2}{n}} \left({z}'\right)+\mathrm{cos}^{\mathrm{2}{n}} \left({z}'\right)}−{I} \\ $$$$\mathrm{2}{I}=\int\:\:\frac{\mathrm{2}\pi\centerdot\mathrm{sin}^{\mathrm{2}{n}} \left({z}'\right)}{\mathrm{sin}^{\mathrm{2}{n}} \left({z}'\right)+\mathrm{cos}^{\mathrm{2}{n}} \left({z}'\right)}\mathrm{d}{z} \\ $$$${I}=\int\:\:\frac{\pi\centerdot\mathrm{sin}^{\mathrm{2}{n}} \left({z}'\right)}{\mathrm{sin}^{\mathrm{2}{n}} +\mathrm{cos}^{\mathrm{2}{n}} \left({z}'\right)}\:\mathrm{d}{z} \\ $$$$\int\:\:\:\frac{\mathrm{sin}^{\mathrm{2}{n}} \left({z}\right)}{\mathrm{sin}^{\mathrm{2}{n}} \left({z}\right)+\mathrm{cos}^{\mathrm{2}{n}} \left({z}\right)}=\int\:\:\:\frac{\mathrm{cos}^{\mathrm{2}{n}} \left({z}\right)}{\mathrm{sin}^{\mathrm{2}{n}} \left({z}\right)+\mathrm{cos}^{\mathrm{2}{n}} \left({z}\right)} \\ $$$$\int\:\:\:\frac{\mathrm{sin}^{\mathrm{2}{n}} \left({z}\right)+\mathrm{cos}^{\mathrm{2}{n}} \left({z}\right)}{\mathrm{sin}^{\mathrm{2}{n}} \left({z}\right)+\mathrm{cos}^{\mathrm{2}{n}} \left({z}\right)}\:\mathrm{d}{z}=\mathrm{2}\pi. \\ $$$$\int\:\:\frac{\mathrm{sin}^{\mathrm{2}{n}} \left({z}\right)}{\mathrm{sin}^{\mathrm{2}{n}} \left({z}\right)+\mathrm{cos}^{\mathrm{2}{n}} \left({z}\right)}\mathrm{d}{z}=\pi \\ $$$$\therefore\pi^{\mathrm{2}} \\ $$

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