Question

${\int }_0^{2\pi }\frac{x{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$.

• A. ${\pi }^2$
• B. $2{\pi }^2$
• C. $4{\pi }^2$
• D. $8{\pi }^2$

Answer 1

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

$I={\int }_0^{2\pi }\frac{x{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$

$\Rightarrow I={\int }_0^{2\pi }\frac{\left(2\pi -x\right){\sin }^{2n}\left(2\pi -x\right)}{{\sin }^{2n}\left(2\pi -x\right)+{\cos }^{2n}\left(2\pi -x\right)}dx$ ($\because {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx$)

$\Rightarrow I={\int }_0^{2\pi }\frac{2\pi {\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx-{\int }_0^{2\pi }\frac{x{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$

$\Rightarrow I={\int }_0^{2\pi }\frac{2\pi {\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx-I$

$\Rightarrow 2I=$${\int }_0^{2\pi }\frac{2\pi {\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$

$\Rightarrow 2I=2\pi {\int }_0^{2\pi }\frac{{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$

$\Rightarrow I=2\pi {\int }_0^{\pi }\frac{{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$ ($\because {\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{2a}f\left(x\right)dx$)

$\Rightarrow I=4\pi {\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$ ---(i) ($\because {\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{2a}f\left(x\right)dx$)

$\Rightarrow I=4\pi {\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{2n}x}{{\cos }^{2n}x+{\sin }^{2n}x}dx$ ---(ii) ($\because {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx$)

Adding (i) and (ii) we get

$\Rightarrow 2I=4\pi {\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2n}x+{\cos }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$

$\Rightarrow I=2\pi {\int }_0^{\frac{\pi }{2}}dx$

$\Rightarrow I=2\pi {\left[x\right]}_0^{\frac{\pi }{2}}$

$\Rightarrow I=2\pi \times \frac{\pi }{2}$

$\Rightarrow I={\pi }^2$