Question

For $n>0,\underset{0}{\overset{2\pi }{\int }}\frac{x{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx=\dots \dots .$

Answer 1

Let $I=\underset{0}{\overset{2\pi }{\int }}\frac{x{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx\dots$ (i)
$I=\underset{0}{\overset{2n}{\int }}\frac{(2\pi -x)[\sin (2\pi -x){]}^{2n}}{[\sin (2\pi -x){]}^{2n}+[\cos (2\pi -x){]}^{2n}}dx$
$\left[\because \underset{0}{\overset{a}{\int }}f(x)dx=\underset{0}{\overset{a}{\int }}f(a-x)dx\right]$
$I=\underset{0}{\overset{2\pi }{\int }}\frac{(2\pi -x)\cdot {\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$
$\Rightarrow I=\underset{0}{\overset{2\pi }{\int }}\frac{2\pi {\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx-$
$\underset{0}{\overset{2\pi }{\int }}\frac{x{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$
$\Rightarrow I=\underset{0}{\overset{2\pi }{\int }}\frac{2\pi {\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx-1$
[from Eq.(i)]
$\Rightarrow I=\underset{0}{\overset{2\pi }{\int }}\frac{\pi {\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$
$\Rightarrow I=\pi [\underset{0}{\overset{\pi }{\int }}\frac{\pi {\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$
$+\underset{0}{\overset{\pi }{\int }}\frac{{\sin }^{2n}(2\pi -x)}{{\sin }^{2n}(2\pi -x)+{\cos }^{2n}(2\pi -x)}dx]$
$\left[\text{ using property }\underset{0}{\overset{2a}{\int }}f(x)dx=\underset{0}{\overset{a}{\int }}[f(x)+f(2a-x)]dx\right]$
$=\pi [\underset{0}{\overset{\pi }{\int }}\frac{{\sin }^{2n}xdx}{{\sin }^{2n}x+{\cos }^{2n}x}dx$
$+\underset{0}{\overset{\pi }{\int }}\frac{{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx]$
$\Rightarrow I=2\pi \underset{0}{\overset{\pi }{\int }}\frac{{\sin }^{2n}xdx}{{\sin }^{2n}x+{\cos }^{2n}x}dx]$
$\Rightarrow I=4\pi \left[\underset{0}{\overset{\pi /2}{\int }}\frac{{\sin }^{2n}xdx}{{\sin }^{2n}+{\cos }^{2n}x}dx\right]$ ....(ii)
$\Rightarrow I=4\pi \underset{0}{\overset{\pi /2}{\int }}\frac{{\sin }^{2n}(\pi /2-x)}{{\sin }^{2n}(\pi /2-x)+{\cos }^{2n}(\pi /2-x)}dx$
$\Rightarrow I=4\pi \underset{0}{\overset{\pi /2}{\int }}\frac{{\cos }^{2n}x}{{\cos }^{2n}x+{\sin }^{2n}x}dx\dots$ (iii)
On adding Eqs. (ii) and (iii), we get
$2I=4\pi \underset{0}{\overset{\pi /2}{\int }}\frac{{\sin }^{2n}x+{\cos }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$
$\Rightarrow 2I=4\pi \underset{0}{\overset{\pi /2}{\int }}1dx=4\pi [x{]}_0^{\pi /2}=4\pi \cdot \pi /2$
$\Rightarrow I={\pi }^2$