Question

The value of $\underset{-2021\pi }{\overset{2021\pi }{\int }}\left\{x^{2021}(1+\tan (\frac{\pi }{3}-x))(1+\tan (x-\frac{\pi }{12}))\right\}dx$ is

• A. $\frac{\pi }{2021}$
• B. $\frac{\pi }{2022}$
• C. $0$
• D. $\frac{(2022{)}^2}{2}$

Answer 1

The correct option is C $0$

As we know that, $(1+\tan A)(1+\tan B)=2$
when $A+B=\frac{\pi }{4}$
Here, $A=\frac{\pi }{3}-x,B=x-\frac{\pi }{12}$
$\Rightarrow A+B=\frac{\pi }{3}-x+x-\frac{\pi }{12}=\frac{4\pi -\pi }{12}=\frac{3\pi }{12}=\frac{\pi }{4}$
$\therefore \underset{-2021\pi }{\overset{2021\pi }{\int }}\left\{x^{2021}(1+\tan (\frac{\pi }{3}-x))(1+\tan (x-\frac{\pi }{12}))\right\}dx$
$=\underset{-2021\pi }{\overset{2021\pi }{\int }}{x}^{2021}\left(2\right)dx$
We know that,
$\underset{-a}{\overset{a}{\int }}f\left(x\right)dx=\{\begin{array}{c}2\underset{0}{\overset{a}{\int }}f\left(x\right)dx,\text{if}f\left(x\right)\text{is even}\\ 0,\text{if}f\left(x\right)\text{is odd}\end{array}$
Here, $f\left(x\right)=2\cdot x^{2021}$
$\Rightarrow f(-x)=-2x^{2021}$
$\Rightarrow f(-x)=-f\left(x\right)$, thus $f\left(x\right)$ is odd.
$\Rightarrow \underset{-2021\pi }{\overset{2021\pi }{\int }}2\cdot x^{2021}dx=0$