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Question: For \(n > 0,\int_{0}^{2\pi}{\frac{x\sin^{2n}x}{\sin^{2n}x + \cos^{2n}x}dx}\) is equal to...

For n>0,∫02πxsin⁡2nxsin⁡2nx+cos⁡2nxdxn > 0,\int_{0}^{2\pi}{\frac{x\sin^{2n}x}{\sin^{2n}x + \cos^{2n}x}dx} is equal to

A

π2\pi^{2}

B

2π22\pi^{2}

C

3π23\pi^{2}

D

4π24\pi^{2}

Answer

π2\pi^{2}

Explanation

Solution

I=∫02πxsin⁡2nxdxsin⁡2nx+cos⁡2nxI = \int_{0}^{2\pi}\frac{x\sin^{2n}xdx}{\sin^{2n}x + \cos^{2n}x} and

I=∫02π(2π−x)sin⁡2n(2π−x)dxsin⁡2n(2π−x)+cos⁡2n(2π−x)I = \int_{0}^{2\pi}\frac{(2\pi - x)\sin^{2n}(2\pi - x)dx}{\sin^{2n}(2\pi - x) + \cos^{2n}(2\pi - x)} [∵∫0af(x)=∫0af(a−x)]\left\lbrack \because\int_{0}^{a}{f(x) = \int_{0}^{a}{f(a - x)}} \right\rbrack

∴ 2I=2π∫02πsin⁡2nπsin⁡2nx+cos⁡2nxdx2I = 2\pi\int_{0}^{2\pi}{\frac{\sin^{2n}\pi}{\sin^{2n}x + \cos^{2n}x}dx}

⇒ I=π∫02πsin⁡2nxsin⁡2nx+cos⁡2nxdxI = \pi\int_{0}^{2\pi}\frac{\sin^{2n}x}{\sin^{2n}x + \cos^{2n}x}dxusing ∫0nTf(x)=n∫0Tf(x)dx\int_{0}^{nT}{f(x) = n\int_{0}^{T}{f(x)dx}}

∴ I=4π∫0π⥂/⥂2sin⁡2nxsin⁡2nx+cos⁡2nxdxI = 4\pi\int_{0}^{\pi ⥂ / ⥂ 2}{\frac{\sin^{2n}x}{\sin^{2n}x + \cos^{2n}x}dx} ⇒ I=4π(π⥂/⥂4)=π2I = 4\pi(\pi ⥂ / ⥂ 4) = \pi^{2}.