What Is the Best Approach to Integrate ln(sec θ - tan θ) from -π/4 to π/4?

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SUMMARY

The integral I = ∫-π/4π/4 ln(sec θ - tan θ) dθ can be evaluated using properties of definite integrals. By recognizing that the integrand can be expressed as I = ∫-π/4π/4 ln(sec θ + tan θ) dθ, the integral can be simplified. The function f(θ) = ln(sec θ - tan θ) is identified as an odd function, leading to the conclusion that the integral evaluates to zero since f(θ) + f(-θ) = 0.

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PhysicoRaj
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Homework Statement



Integrate:I=\int_{-π/4}^{π/4} \ln{(\sec θ-\tan θ)}\,dθ

Homework Equations



Properties of definite integrals, basic integration formulae, trigonometric identities.

The Attempt at a Solution


By properties of definite integrals, the same integral I wrote as equivalent toI=\int_{-π/4}^{π/4} \ln{(\sec θ+\tan θ)}\,dθ.
Because\int_{a}^{b} f(x)\,dx=\int_{a}^{b} f(a+b-x)\,dx(replacing θ by π/4-π/4-θ) Now, I think of adding these two integrals to form an equation and solving for I but I'm messing up. Am I doing wrong? Is there any better/easy way?
Thanks for your time and help.
 
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PhysicoRaj said:

Homework Statement



Integrate:I=\int_{-π/4}^{π/4} \ln{(\sec θ-\tan θ)}\,dθ

Homework Equations



Properties of definite integrals, basic integration formulae, trigonometric identities.

The Attempt at a Solution


By properties of definite integrals, the same integral I wrote as equivalent toI=\int_{-π/4}^{π/4} \ln{(\sec θ+\tan θ)}\,dθ.
Because\int_{a}^{b} f(x)\,dx=\int_{a}^{b} f(a+b-x)\,dx(replacing θ by π/4-π/4-θ) Now, I think of adding these two integrals to form an equation and solving for I but I'm messing up. Am I doing wrong? Is there any better/easy way?
Thanks for your time and help.

What do you get if you add them? It's quite obvious.

Instead, you can notice that the integrand is an odd function, do you see why?
 
I tried that at first but I couldn't come to a decision, I can't see if it's odd or even. The secant is an even function right?
 
PhysicoRaj said:
I tried that at first but I couldn't come to a decision, I can't see if it's odd or even. The secant is an even function right?

Yes, secant is an even function.

Define ##f(\theta)=\ln\left(\sec\theta-\tan\theta\right)##. Hence, ##f(-\theta)=\ln\left(\sec\theta+\tan\theta\right)##. What do you get if you add ##f(\theta)## and ##f(-\theta)##? Please show the attempt.
 
It's ln(1)=0
 
PhysicoRaj said:
It's ln(1)=0

Right!

So ##f(\theta)+f(-\theta)=0##. Do you see why it is an odd function? :)
 
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Ahh.. I get it right now Pranav, thanks a lot! :smile:
 

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