Integral of tan^2 ln^2: $\pi/2 (1-\ln 2)$

[Tony1]

Show that,

$$\int_{0}^{\pi/2}{\ln^2 (\tan^2 \theta)\over \pi^2+\ln^2 (\tan^2 \theta)}\mathrm d\theta=\color{red}{\pi\over 2}\color{green}(1-\ln 2)$$

 

[jvicens]

Hello everyone,

I came across this interesting integral while studying for my calculus exam and I was wondering if anyone could help me solve it. I have tried using substitution, integration by parts, and trigonometric identities, but I can't seem to get the right answer. Any help or guidance would be greatly appreciated. Thank you!
I will be happy to help you solve this integral. First, let's rewrite the integral using the property of logarithms: $\ln^2 (\tan^2 \theta)=2\ln(\tan \theta)$. This gives us:

$$\int_{0}^{\pi/2}{2\ln(\tan \theta)\over \pi^2+\ln^2 (\tan^2 \theta)}\mathrm d\theta$$

Next, we can use the substitution $u=\tan \theta$, which gives us $\mathrm d\theta={\mathrm du\over \cos^2 \theta}$. Substituting this into our integral, we get:

$$\int_{0}^{\pi/2}{2\ln u\over \pi^2+\ln^2 u}\cos^2 \theta\mathrm du$$

Using the trigonometric identity $\cos^2 \theta={1\over 1+\tan^2 \theta}$, we can rewrite the integral as:

$$\int_{0}^{\pi/2}{2\ln u\over \pi^2+\ln^2 u}{1\over 1+\tan^2 \theta}\mathrm du$$

Now, let's use another substitution, $v=\ln u$, which gives us $\mathrm du={\mathrm dv\over u}$. Substituting this into our integral, we get:

$$\int_{-\infty}^{\infty}{2v\over \pi^2+v^2}{1\over 1+e^{2v}}\mathrm dv$$

This integral can be solved using the method of partial fractions. After solving for the coefficients, we get:

$$\int_{-\infty}^{\infty}{2v\over \pi^2+v^2}{1\over 1+e^{2v}}\mathrm dv={1\over 2}\left({\pi\over 2}+\ln 2\right)$$

Substituting back in our original variables,