Evaluate ∫−π3π log⁡(sec⁡θ−tan⁡θ)dθ .

Let I=∫−π3π log⁡(sec⁡θ−tan⁡θ)dθ----------(1) Using the property ∫ab f(x)dx=∫ab f(a+b−x)dx, we get I=∫−π3π log⁡[sec⁡(2π−θ)−tan⁡(2π−θ)]dθ=∫−π3π log⁡[sec⁡θ+tan⁡θ]dθ------(2) Adding equations (1) and (2), we get 2I=∫−π3π log⁡[(sec⁡θ−tan⁡θ)(sec⁡θ+tan⁡θ)]dθ   =∫−π3π log⁡(1)dθ=∫−π3π 0.dθ=0⇒I=0