Question Number 200684 by Spillover last updated on 21/Nov/23
$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int_{−\mathrm{4}\pi} ^{\mathrm{4}\pi} \:\:\:\frac{\mid{x}\mid\:\mathrm{sin}\:^{\mathrm{2}{n}} {x}}{\mathrm{sin}\:^{\mathrm{2}{n}} {x}+\mathrm{cos}\:^{\mathrm{2}{n}} {x}}{dx} \\ $$$$ \\ $$$$ \\ $$
Answered by witcher3 last updated on 22/Nov/23
![=2∫_0 ^(4π) ((xsin^(2n) (x))/(cos^(2n) (x)+sin^(2n) (x)))dx =2[∫_0 ^(2π) ((xsin^(2n) (x))/(sin^(2n) (x)+cos^(2n) (x))) +(((2π+x)sin^(2n) (x))/(sin^(2n) (x)+cos^(2n) (x)))dx] =4∫_0 ^(2π) (((π+x)sin^(2n) (x))/(sin^(2n) (x)+cos^(2n) (x)))dx =4∫_0 ^π ((π+x)/(sin^(2n) (x)+cos^(2n) (x)))sin^(2n) (x)+4∫_0 ^π ((π+2π−x)/(sin^(2n) (x)+cos^(2n) (x)))sin^(2n) (x) =4∫_0 ^π ((4π)/(sin^(2n) (x)+cos^(2n) (x)))sin^(2n) (x) =16π[∫_0 ^(π/2) ((sin^(2n) (x))/(cos^(2n) (x)+cos^(2n) (x)))+∫_(π/2) ^π ((sin^(2n) (x))/(cos^(2n) (x)+sin^(2n) (x)))dx] x→(π/2)+x in 2nd cos^(2n) (x+(π/2))=sin^(2n) (x);sin^(2n) (x+(π/2))=cos^(2n) (x) I=16π∫_0 ^(π/z) ((sin^(2n) (x)+cos^(2n) (x))/(cos^(2n) (x)+sin^(2n) (x)))dx =16π∫_0 ^(π/2) dx=8π^2](https://www.tinkutara.com/question/Q200714.png)
$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{4}\pi} \frac{\mathrm{xsin}^{\mathrm{2n}} \left(\mathrm{x}\right)}{\mathrm{cos}^{\mathrm{2n}} \left(\mathrm{x}\right)+\mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}\right)}\mathrm{dx} \\ $$$$=\mathrm{2}\left[\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{\mathrm{xsin}^{\mathrm{2n}} \left(\mathrm{x}\right)}{\mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}\right)+\mathrm{cos}^{\mathrm{2n}} \left(\mathrm{x}\right)}\:+\frac{\left(\mathrm{2}\pi+\mathrm{x}\right)\mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}\right)}{\mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}\right)+\mathrm{cos}^{\mathrm{2n}} \left(\mathrm{x}\right)}\mathrm{dx}\right] \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{\left(\pi+\mathrm{x}\right)\mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}\right)}{\mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}\right)+\mathrm{cos}^{\mathrm{2n}} \left(\mathrm{x}\right)}\mathrm{dx} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\pi} \frac{\pi+\mathrm{x}}{\mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}\right)+\mathrm{cos}^{\mathrm{2n}} \left(\mathrm{x}\right)}\mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}\right)+\mathrm{4}\int_{\mathrm{0}} ^{\pi} \frac{\pi+\mathrm{2}\pi−\mathrm{x}}{\mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}\right)+\mathrm{cos}^{\mathrm{2n}} \left(\mathrm{x}\right)}\mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}\right) \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{4}\pi}{\mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}\right)+\mathrm{cos}^{\mathrm{2n}} \left(\mathrm{x}\right)}\mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}\right) \\ $$$$=\mathrm{16}\pi\left[\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}\right)}{\mathrm{cos}^{\mathrm{2n}} \left(\mathrm{x}\right)+\mathrm{cos}^{\mathrm{2n}} \left(\mathrm{x}\right)}+\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \frac{\mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}\right)}{\mathrm{cos}^{\mathrm{2n}} \left(\mathrm{x}\right)+\mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}\right)}\mathrm{dx}\right] \\ $$$$\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}+\mathrm{x}\:\mathrm{in}\:\mathrm{2nd} \\ $$$$\mathrm{cos}^{\mathrm{2n}} \left(\mathrm{x}+\frac{\pi}{\mathrm{2}}\right)=\mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}\right);\mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}+\frac{\pi}{\mathrm{2}}\right)=\mathrm{cos}^{\mathrm{2n}} \left(\mathrm{x}\right) \\ $$$$\mathrm{I}=\mathrm{16}\pi\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{z}}} \frac{\mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}\right)+\mathrm{cos}^{\mathrm{2n}} \left(\mathrm{x}\right)}{\mathrm{cos}^{\mathrm{2n}} \left(\mathrm{x}\right)+\mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}\right)}\mathrm{dx} \\ $$$$=\mathrm{16}\pi\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{dx}=\mathrm{8}\pi^{\mathrm{2}} \\ $$