Evaluate the following integral: ∫log(sec x + tan x)dx, x ∈ [0, 2π]

Question

Evaluate the following integral:

\(\int\limits_{0}^{2\pi} \)log(sec x + tan x)dx

Answer 1

Let us assume

I =  \(\int\limits_{0}^{2\pi} \)log(sec x + tan x)dx.....equation 1

By property, we know that \(\int\limits_{a}^{b} \)f(x)dx = \(\int\limits_{a}^{b} \)(a + b - c) dx

We know that sec (2π – x) = sec (x)

tan(2π – x)=–tan (x)

thus I = \(\int\limits_{0}^{2\pi} \)log(sec(x) - tan(x))dx.....equation 2

Adding equations 1 and equation 2, we get,

2I =  \(\int\limits_{0}^{2\pi} \)log(sec x + tan x)dx +  \(\int\limits_{0}^{2\pi} \)log(sec(x) - tan(x))dx

We know log (a) + log (b) = log(ab)

We know Trigonometric identity sec² x – tan² x = 1

We know

where b is the upper limit and a is lower and f(x) is integral function

2I = \([0]_0^{2\pi}\)

Thus I = 0.