Answer:
Let \[l=\int_{0}^{2\pi }{\frac{x{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}\,dx}\] ?(i) Then, \[l=\int_{0}^{2\pi }{\frac{(2\pi -x){{\sin }^{2n}}(2\pi -x)}{{{\sin }^{2n}}(2\pi -x)+{{\cos }^{2n}}(2\pi -x)}\,dx}\] \[\left[ \because \,\,\,\int_{0}^{a}{f(x)\,dx}=\int_{0}^{a}{f(a-x)\,dx} \right]\] \[=\int_{0}^{2\pi }{\frac{(2\pi -x){{\sin }^{2n}}}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}\,dx}\] ?(ii) \[[\because sin(2\pi -x)=-\sin x,\cos (2\pi -x)=\cos \,x]\] On adding Eqs. (i) and (ii), we get \[2l=\int_{0}^{2\pi }{\frac{x{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}\,dx}\] \[+\int_{0}^{2\pi }{\frac{(2\pi -x){{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}\,dx}\] \[=\int_{0}^{2\pi }{\frac{2\pi {{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}\,dx}\] \[\Rightarrow \] \[l=\int_{0}^{2\pi }{\frac{\pi {{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}\,dx}\] [dividing both sides by 2] \[=2\pi \int_{0}^{\pi }{\frac{{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}\,dx}\] \[\left[ \because \,\,\,\int_{0}^{2a}{f(x)dx=2\int_{0}^{a}{f(x)dx,\,\,\text{if}\,\,f(2a-x)=f(x)}} \right]\] \[\Rightarrow \] \[l=4\pi \int_{0}^{\pi /2}{\frac{{{\sin }^{2n}}x\,dx}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}\,dx}\] ?(iii) \[\left[ \because \,\,\,\int_{0}^{a}{f(x)dx}=2\int_{0}^{a/2}{f(x)dx},\,\,\,\text{if}\,\,\,f(a-x)=f(x) \right]\] \[\Rightarrow \] \[l=4\pi \int_{0}^{\pi /2}{\frac{{{\sin }^{2n}}\left( \frac{\pi }{2}-x \right)}{{{\sin }^{2n}}\left( \frac{\pi }{2}-x \right)+{{\cos }^{2n}}\left( \frac{\pi }{2}-x \right)}\,dx}\] \[\left[ \because \,\,\,\int_{0}^{a}{f(x)\,dx=\int_{0}^{a}{f(a-x)dx}} \right]\] \[\Rightarrow \] \[l=4\pi \int_{0}^{\pi /2}{\frac{{{\cos }^{2n}}x}{{{\cos }^{2n}}x+{{\sin }^{2n}}x}\,dx}\] ?(iv) On adding Eqs. (i) and (ii), we get \[2l=4\pi \int_{0}^{\pi /2}{1\,dx=4\pi [x]_{0}^{\pi /2}=4\pi \cdot \frac{\pi }{2}=2{{\pi }^{2}}}\] \[\therefore \] \[l={{\pi }^{2}}\] Hence proved.
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