Question

For $n>0,{\int }_0^{2\pi }\frac{xsi{n}^{2n}x}{si{n}^{2n}x+co{s}^{2n}x}dx=$

• A. $\frac{\pi }{2}$
• B. $4{\pi }^2$
• C. ${\pi }^2$
• D. $\pi$

Answer 1

The correct option is C ${\pi }^2$

$I={\int }_0^{2\pi }\frac{xsi{n}^{2n}x}{si{n}^{2n}x+co{s}^{2n}x}dx...\left(1\right)\Rightarrow I={\int }_0^{2\pi }\frac{(2\pi -x)si{n}^{2n}(\pi -x)}{si{n}^{2\pi }(2\pi -x)+co{s}^{2n}(2\pi -x)}dx\left[\text{Using}{\int }_0^{a}f\right(x)dx=f(a-x\left)dx\right]I={\int }_0^{2\pi }\frac{(2\pi -x)si{n}^{2n}x}{si{n}^{2n}x+co{s}^{2n}x}...\left(2\right)$
Adding (1) and (2) we get
$2I={\int }_0^{2\pi }\frac{2\pi si{n}^{2n}x}{si{n}^{2n}x+co{s}^{2n}x}dx\Rightarrow I=\pi {\int }_0^{2\pi }\frac{si{n}^{2n}x}{si{n}^{2n}x+co{s}^{2n}x}dx\Rightarrow I=2\pi {\int }_0^{\pi }\frac{si{n}^{2n}x}{si{n}^{2n}x+co{s}^{2n}x}dx\left[\text{Using}{\int }_0^{2a}f\right(x)dx=2{\int }_0^{a}f(x\left)dx\text{if}f\right(2a-x)=f(x\left)\right]\Rightarrow I=4\pi {\int }_0^{\frac{\pi }{2}}\frac{si{n}^{2n}x}{si{n}^{2n}x+co{s}^{2n}x}dx...\left(3\right)$
[Using above property again]
$\Rightarrow I=4\pi {\int }_0^{\frac{\pi }{2}}\frac{co{s}^{2n}x}{co{s}^{2n}x+si{n}^{2n}x}dx...\left(4\right)$
$\left[\text{Using}{\int }_0^{a}f\right(x)dx=f(a-x\left)dx\right]$
Adding (3) and (4) we get
$2I=4\pi {\int }_0^{\frac{\pi }{2}}1.dx=4\pi (\frac{\pi }{2}-0)=2{\pi }^2\Rightarrow I={\pi }^2$