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

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

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$Evaluate \int_{- π}^{3 π} \log (\sec θ - \tan θ) d θ .$

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Detailed Solution

$Let I = \int_{- π}^{3 π} \log (\sec θ - \tan θ) d θ$ ----------(1)

$Using the property \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x, we get$

$I = \int_{- π}^{3 π} \log [\sec (2 π - θ) - \tan (2 π - θ)] d θ$

$= \int_{- π}^{3 π} \log [\sec θ + \tan θ] d θ - - - - - - (2)$

$Adding equations (1) and (2), we get$

$\begin{array}{l} 2 I = \int_{- π}^{3 π} \log [(\sec θ - \tan θ) (\sec θ + \tan θ)] d θ \\ = \int_{- π}^{3 π} \log (1) d θ = \int_{- π}^{3 π} 0 . d θ = 0 \\ \Rightarrow I = 0 \end{array}$