How to Prove the Integral Result Equals ln(2) - ζ(2)/4?

[Tony1]

How to show that,

$$8\pi\int_{0}^{\pi/2}\cos^2x\cdot{\ln^2(\tan^2 x)\over [\pi^2+\ln^2(\tan^2 x)]^2}\mathrm dx=\ln 2-{1\over 4}\zeta(2)$$

 

[jvicens]

To show this, we can use the properties of logarithms and trigonometric identities to simplify the integrand.

First, we can rewrite the integrand as:

$$\cos^2x\cdot{\ln^2(\tan^2 x)\over [\pi^2+\ln^2(\tan^2 x)]^2} = {\cos^2x\cdot\ln^2(\sin^2x)\over [\pi^2+\ln^2(\sin^2x)]^2}$$

Next, we can use the identity $\ln(\sin^2x) = 2\ln(\sin x)$ to simplify the expression further:

$$ {\cos^2x\cdot\ln^2(\sin^2x)\over [\pi^2+\ln^2(\sin^2x)]^2} = {\cos^2x\cdot[2\ln(\sin x)]^2\over [\pi^2+[2\ln(\sin x)]^2]^2}$$

Now, we can use the double angle formula for cosine to rewrite $\cos^2x$ as $(1+\cos2x)/2$:

$${\cos^2x\cdot[2\ln(\sin x)]^2\over [\pi^2+[2\ln(\sin x)]^2]^2} = {(1+\cos2x)\cdot[2\ln(\sin x)]^2\over [\pi^2+[2\ln(\sin x)]^2]^2}$$

Using the identity $\ln(\sin x) = -\ln(\cos x)$, we can rewrite the expression as:

$$ {(1+\cos2x)\cdot[2\ln(\cos x)]^2\over [\pi^2+[2\ln(\cos x)]^2]^2}$$

Now, we can use the substitution $u = \cos x$ to change the limits of integration and rewrite the integral as:

$$8\pi\int_{0}^{1}{(1+u)\cdot[2\ln u]^2\over [\pi^2+[2\ln u]^2]^2}\mathrm du$$

Using the power rule for logarithms, we can simplify the expression further:

$$8\pi\int_{0}^{1}{(1+u)\cdot[2\ln u]^2\over