For n 0, int_0^{2pi } {frac{{x{{sin }^{2n}} | vedclass

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(D) Let $I = \int_0^{2\pi } {\frac{{x{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $.Using the property $\int_0^a {f(x)dx = \int_0^a {f(a

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${\pi ^2}$

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(D) Let $I = \int_0^{2\pi } {\frac{{x{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $.Using the property $\int_0^a {f(x)dx = \int_0^a {f(a - x)dx} }$,we have:$I = \int_0^{2\pi } {\frac{{(2\pi - x){{\sin }^{2n}}(2\pi - x)}}{{{{\sin }^{2n}}(2\pi - x) + {{\cos }^{2n}}(2\pi - x)}}dx} $$I = \int_0^{2\pi } {\frac{{(2\pi - x){{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $Adding the two expressions for $I$:$2I = \int_0^{2\pi } {\frac{{2\pi {{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $$I = \pi \int_0^{2\pi } {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $Using the property $\int_0^{2a} {f(x)dx = 2\int_0^a {f(x)dx} }$ if $f(2a-x) = f(x)$:$I = 2\pi \int_0^{\pi } {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $$I = 4\pi \int_0^{\pi /2} {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $ (Equation $i$)Similarly,$I = 4\pi \int_0^{\pi /2} {\frac{{{{\cos }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $ (Equation $ii$)Adding $(i)$ and $(ii)$:$2I = 4\pi \int_0^{\pi /2} {1 dx} = 4\pi \left( \frac{\pi }{2} \right) = 2{\pi ^2}$Therefore,$I = {\pi ^2}$.

For $n > 0,$ $\int_0^{2\pi } {\frac{{x{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}\,dx = } $

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