Substitute y = tan(x) and dy = sec²(x) dx to transform the integral to
[tex]\displaystyle \int_0^{\frac\pi2} \cot(x) \ln(\sec(x)) \, dx = \frac12 \int_0^\infty \frac{\ln(1+y^2)}{y(1+y^2)} \, dy[/tex]
and we split the integral at y = 1.
Examining the one over [0, 1], expand into partial fractions :
[tex]\dfrac1{y(1+y^2)} = \dfrac1y - \dfrac y{1+y^2}[/tex]
Then
[tex]\displaystyle \int_0^1 \frac{\ln(1+y^2)}{y(1+y^2)} \, dy = \int_0^1 \frac{\ln(1+y^2)}y \, dy - \int_0^1 \frac{y \ln(1+y^2)}{1 + y^2} \, dy[/tex]
In the integral over [1, ∞), substitute y = 1/z :
[tex]\displaystyle \int_1^\infty \frac{\ln(1+y^2)}{y(1+y^2)} \, dy = \int_0^1 \frac{z \ln\left(1+\frac1{z^2}\right)}{1+z^2} \, dz \\\\ = \int_0^1 \frac{z \ln\left(1+z^2\right)}{1+z^2} \, dz - 2 \int_0^1 \frac{z \ln(z)}{1+z^2} \, dz[/tex]
Some terms cancel and we're left with
[tex]\displaystyle \int_0^\infty \frac{\ln(1+y^2)}{y(1+y^2)} \, dy = \frac12 \int_0^1 \frac{\ln(1+y^2)}y \, dy - \int_0^1 \frac{y \ln(y)}{1+y^2} \, dy[/tex]
Use the series expansions of ln(1 + y) and 1/(1 - y) - both are valid for |y| < 1.
[tex]\displaystyle \int_0^1 \frac{\ln(1+y^2)}y \, dy = - \sum_{n=1}^\infty \frac{(-1)^n}n \int_0^1 y^{2n-1} \, dy = -\frac12 \sum_{n=1}^\infty \frac{(-1)^n}{n^2} = \frac{\pi^2}{24}[/tex]
[tex]\displaystyle \int_0^1 \frac{y \ln(y)}{1+y^2} \, dy = \sum_{n=0}^\infty (-1)^n \int_0^1 y^{2n+1} \ln(y) \, dy = -\frac14 \sum_{n=0}^\infty \frac{(-1)^n}{(n+1)^2} = -\frac{\pi^2}{48}[/tex]
Putting everything together, we have
[tex]\displaystyle \int_0^{\frac\pi2} \cot(x) \ln(\sec(x)) \, dx = \frac12\times\frac{\pi^2}{24} - \left(- \frac{\pi^2}{48}\right) = \boxed{\frac{\pi^2}{24}}[/tex]
