Question

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

• A. ${3\pi }^2$
• B. ${5\pi }^2$
• C. ${\pi }^2$
• D. $Noneofthese$

Answer 1

Answer: (C)

${\pi }^2$

Consider the given integral.

$I=\underset{0}{\overset{2\pi }{\int }}\frac{x{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$ …… (1)

Use the property,

$\underset{0}{\overset{a}{\int }}f\left(t\right)dt=\underset{0}{\overset{a}{\int }}f(a-t)dt$

Therefore,

$I=\underset{0}{\overset{2\pi }{\int }}\frac{(2\pi -x){\sin }^{2n}(2\pi -x)}{{\sin }^{2n}(2\pi -x)+{\cos }^{2n}(2\pi -x)}dx$

$I=\underset{0}{\overset{2\pi }{\int }}\frac{(2\pi -x){\sin }^{2n}\left(x\right)}{{\sin }^{2n}\left(x\right)+{\cos }^{2n}\left(x\right)}dx$ …… (2)

Add equations (1) and (2).

$2I=\underset{0}{\overset{2\pi }{\int }}\frac{(2\pi -x+x){\sin }^{2n}\left(x\right)}{{\sin }^{2n}\left(x\right)+{\cos }^{2n}\left(x\right)}dx$

$2I=2\pi \underset{0}{\overset{2\pi }{\int }}\frac{{\sin }^{2n}\left(x\right)}{{\sin }^{2n}\left(x\right)+{\cos }^{2n}\left(x\right)}dx$

$I=\pi \underset{0}{\overset{2\pi }{\int }}\frac{{\sin }^{2n}\left(x\right)}{{\sin }^{2n}\left(x\right)+{\cos }^{2n}\left(x\right)}dx$

Again, using property $\underset{0}{\overset{2a}{\int }}f\left(t\right)dt=2\underset{0}{\overset{a}{\int }}f\left(t\right)dt$, if $f(2a-t)=f\left(t\right)$, we have

$I=2\pi \underset{0}{\overset{\pi }{\int }}\frac{{\sin }^{2n}\left(x\right)}{{\sin }^{2n}\left(x\right)+{\cos }^{2n}\left(x\right)}dx$

Again, use same property.

$I=4\pi \underset{0}{\overset{\pi /2}{\int }}\frac{{\sin }^{2n}\left(x\right)}{{\sin }^{2n}\left(x\right)+{\cos }^{2n}\left(x\right)}dx$ …… (3)

$I=4\pi \underset{0}{\overset{\pi /2}{\int }}\frac{{\sin }^{2n}(\frac{\pi }{2}-x)}{{\sin }^{2n}(\frac{\pi }{2}-x)+{\cos }^{2n}(\frac{\pi }{2}-x)}dx$

$I=4\pi \underset{0}{\overset{\pi /2}{\int }}\frac{{\cos }^{2n}\left(x\right)}{{\sin }^{2n}\left(x\right)+{\cos }^{2n}\left(x\right)}dx$ …… (4)

Add equations (3) and (4).

$2I=4\pi \underset{0}{\overset{\pi /2}{\int }}\frac{{\sin }^{2n}\left(x\right)+{\cos }^{2n}\left(x\right)}{{\sin }^{2n}\left(x\right)+{\cos }^{2n}\left(x\right)}dx$

$2I=4\pi \underset{0}{\overset{\pi /2}{\int }}1dx$

$I=2\pi [\frac{\pi }{2}-0]$

$I={\pi }^2$

Hence, the value of the integral is ${\pi }^2$.