Question

Evaluate each of the following integral:

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

 

Answer 1

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

Then,

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

Adding (1) and (2), we get

$$2I={\int }_0^{2\pi }[\log (\sec x+\tan x)+\log (\sec x-\tan x)]dx$$
$$\Rightarrow 2I={\int }_0^{2\pi }\log ({\sec }^{2}x-{\tan }^{2}x)dx$$
$$\Rightarrow 2I={\int }_0^{2\pi }\log 1dx.................(1+{\tan }^{2}x={\sec }^{2}x)$$
$$\Rightarrow 2I=0......................(\log 1=0)$$
$$\Rightarrow I=0$$