Evaluate of each of the following integrals (1-15): `int_0^(2pi)"log"(secx+tanx) dx`

Explanation

`Let I=int_{0}^{2 pi} log (sec x+tan x) d x`

Then,
`
I=int_{0}^{2 pi} log [sec (2 pi-x)+tan (2 pi-x)] d x ldots ldots ldots ldots ldots[int_{0}^{a} f(x) d x=int_{0}^{a} f(a-x) d x] `
`
=int_{0}^{2 pi} log (sec x-tan x) d x ldots ldots ldots ldots ldots ldots(2) `

Adding (1) { and (2), we get
`
2 I=int_{0}^{2 pi}[log (sec x+tan x)+log (sec x-tan x)] d x `
`
Rightarrow 2 I=int_{0}^{2 pi} log (sec ^{2} x-tan ^{2} x) d x `
`
Rightarrow 2 I=int_{0}^{2 pi} log 1 d x ldots ldots ldots ldots ldots ldots(1+tan ^{2} x=sec ^{2} x) `
`
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