Question

Evaluate each of the following integrals:

${\int }_0^{2\mathrm{\pi }}\log \left(\sec x+\tan x\right)dx$

Answer 1

Let I = ${\int }_0^{2\mathrm{\pi }}\log \left(\sec x+\tan x\right)dx$ .....(1)

Then,

$$\begin{gathered} I={\int }_0^{2\mathrm{\pi }}\log \left[\sec \left(2\mathrm{\pi }-x\right)+\tan \left(2\mathrm{\pi }-x\right)\right]dx\left[{\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx\right] \\ ={\int }_0^{2\mathrm{\pi }}\log \left(\sec x-\tan x\right)dx.....\left(2\right) \end{gathered}$$

Adding (1) and (2), we get

$$\begin{gathered} 2I={\int }_0^{2\mathrm{\pi }}\left[\log \left(\sec x+\tan x\right)+\log \left(\sec x-\tan x\right)\right]dx \\ \Rightarrow 2I={\int }_0^{2\mathrm{\pi }}\log \left({\sec }^{2}x-{\tan }^{2}x\right)dx \\ \Rightarrow 2I={\int }_0^{2\mathrm{\pi }}\log 1dx\left(1+{\tan }^{2}x={\sec }^{2}x\right) \\ \Rightarrow 2I=0\left(\log 1=0\right) \\ \Rightarrow I=0 \end{gathered}$$