question_answer

Prove that \[\int_{0}^{2\pi }{\frac{x{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}\,dx}={{\pi }^{2}}.\]

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.