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

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Homework Help Overview

The problem involves integrating the function I=\int_{-π/4}^{π/4} \ln{(\sec θ-\tan θ)}\,dθ, which falls under the subject area of calculus, specifically dealing with definite integrals and properties of logarithmic and trigonometric functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of properties of definite integrals to relate the original integral to another form involving ln(sec θ + tan θ). There is an exploration of whether adding the two integrals could lead to a solution. Some participants question the nature of the integrand, considering whether it is odd or even.

Discussion Status

The discussion is active, with participants exploring different approaches and clarifying concepts related to odd and even functions. Some guidance has been offered regarding the properties of the functions involved, and there is a recognition of the relationship between f(θ) and f(-θ).

Contextual Notes

Participants are navigating through the implications of the properties of the integrand and its behavior over the specified interval. There is a focus on understanding the characteristics of the functions involved without reaching a definitive conclusion.

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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