Question

If $I=\underset{0}{\overset{2\pi }{\int }}\frac{x}{1+{\tan }^{2n}x}dx,n\in N;$ then which of the folllowing statements is/are true

• A. $I={\pi }^2$
• B. $I=\frac{{\pi }^2}{4}$
• C. $\underset{0}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f(a-x)dx\text{can be used to evaluate it}$
• D. $\text{Integrand is an odd function}$

Answer 1

Answer: (A), (C), (D)

$\text{Integrand is an odd function}$

$I=\underset{0}{\overset{2\pi }{\int }}\frac{x}{1+{\tan }^{2n}x}dx\Rightarrow I=\underset{0}{\overset{2\pi }{\int }}\frac{x{\cos }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx\cdots \left(i\right)\Rightarrow I=\underset{0}{\overset{2\pi }{\int }}\frac{(2\pi -x){\cos }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx\cdots \left(ii\right)$
Adding $\left(i\right)$ and $\left(ii\right)$
$2I=2\pi \underset{0}{\overset{2\pi }{\int }}\frac{{\cos }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx$
$\Rightarrow I=2\pi \underset{0}{\overset{\pi }{\int }}\frac{{\cos }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx\{\because \underset{0}{\overset{2a}{\int }}f\left(x\right)dx=2\underset{0}{\overset{a}{\int }}f\left(x\right)dx,\text{when}f\left(x\right)=f(2a-x)\}$
Again using the same property on $I$
$\Rightarrow I=4\pi \underset{0}{\overset{\pi /2}{\int }}\frac{{\cos }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx\cdots \left(iii\right)\left[\text{using}\underset{a}{\overset{b}{\int }}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx\right]\Rightarrow I=4\pi \underset{0}{\overset{\pi /2}{\int }}\frac{{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx\cdots \left(iv\right)$
Adding $\left(iii\right)$ and $\left(iv\right)$
$\Rightarrow 2I=4\pi \cdot \frac{\pi }{2}\Rightarrow I={\pi }^2$