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

[PhysicoRaj]

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 to$$I=\int_{-π/4}^{π/4} \ln{(\sec θ+\tan θ)}\,dθ$$.
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.

 

[Saitama]

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 to$$I=\int_{-π/4}^{π/4} \ln{(\sec θ+\tan θ)}\,dθ$$.
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?

 

[PhysicoRaj]

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?

 

[Saitama]

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.

 

[PhysicoRaj]

It's $ln(1)=0$

 

[Saitama]

It's $ln(1)=0$

Right!

So $f(\theta)+f(-\theta)=0$. Do you see why it is an odd function? :)

 

[PhysicoRaj]

Ahh.. I get it right now Pranav, thanks a lot!