If $\underset{0}{\overset{\pi }{\int }}f\left(\tan x\right)dx=\lambda then\underset{0}{\overset{2\pi }{\int }}f\left(\tan x\right)dx=$

$\begin{array}{l}\underset{0}{\overset{\pi }{\int }}f\left(\tan x\right)dx=\lambda \\ Wecanwrite\underset{0}{\overset{2\pi }{\int }}f\left(\tan x\right)dx\\ =\underset{0}{\overset{\pi }{\int }}f\left(\tan x\right)dx+\underset{\pi }{\overset{2\pi }{\int }}f\left(\tan x\right)dx\\ =\lambda +\underset{0}{\overset{\pi }{\int }}f\left(\tan \right(\pi +x\left)\right)dx\\ \left[\therefore \underset{a+c}{\overset{b+c}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(c+x)dx\right]\\ =\lambda +\underset{0}{\overset{\mathrm{\pi }}{\int }}f\left(\tan x\right)dx\\ =\lambda +\lambda \\ =2\lambda \end{array}$