Evaluate : $\int\limits_0^{\pi/2} \cot x \ln(\sec x) dx$ I want to solve this integral :
$$\int_0^{\pi/2} \cot x \ln(\sec x) dx$$
I tried the following substitution : $\ln(\sec x)=t$ which means $dt=\tan x dx$
$$I=\int_0^\infty \frac{\cot x}{\tan x}tdt=\int_0^\infty \frac{t}{\tan^2 x}dt$$
I'm really disturbed by the $\tan^2 x$, I tried also to substitute $\sec x =t$ but it's not helpful either. Any helpful approach to solve this problem ?
 A: Substitute $t= \tan^2 x$
\begin{align}
\int_0^{\pi/2} \cot x \ln(\sec x) dx= & \>\frac14\int_0^1\frac{\ln(1+t)}{t(1+t)}dt
 + \frac14\int_1^\infty \underset{t\to1/t}{\frac{\ln(1+t)}{t(1+t)}dt}\\
=&\>\frac14 \int_0^1 \frac{\ln(1+t)}{t}dt-\frac14 \int_0^1  \frac{\ln t}{1+t}dt \\
=& \>\frac12 \int_0^1 \frac{\ln (1+t)}{t}dt
=\frac12\cdot \frac{\pi^2}{12}=\frac{\pi^2}{24}
\end{align}
where $\int_0^1 \frac{\ln (1+t)}{t}dt =\frac{\pi^2}{12}$
A: General case
After $u=\sec x$ the integral becomes
$$\int\dfrac{\ln\left(u\right)}{\left(u-1\right)u\left(u+1\right)}\,\mathrm{d}u.$$
With partial fraction decomposition:
$$=\class{steps-node}{\cssId{steps-node-3}{\dfrac{1}{2}}}{\displaystyle\int}\dfrac{\ln\left(u\right)}{u+1}\,\mathrm{d}u-{\displaystyle\int}\dfrac{\ln\left(u\right)}{u}\,\mathrm{d}u+\class{steps-node}{\cssId{steps-node-4}{\dfrac{1}{2}}}{\displaystyle\int}\dfrac{\ln\left(u\right)}{u-1}\,\mathrm{d}u.$$
After some trivial substitutions and two integration by parts we can see that this is equal to
$$\dfrac{\ln\left(u\right)\ln\left(u+1\right)}{2}-\dfrac{\ln^2\left(u\right)}{2}+\dfrac{\operatorname{Li}_2\left(-u\right)}{2}-\dfrac{\operatorname{Li}_2\left(1-u\right)}{2}.$$
Boundaries $0$ and $\pi/2$
To finish the general case you only have to substitute back (and add the constant). I will not do this now because the expression will get unnecessarily messy.
Remember we set $u=\sec x$, which means that $0\to 1$ and $\pi/2^- \to \infty$. Final solution is (please forgive me for the use of "$\infty$", will add the limits later on)
$$\left[\dfrac{\ln\left(\infty\right)\ln\left(\infty+1\right)}{2}-\dfrac{\ln^2\left(\infty\right)}{2}+\dfrac{\operatorname{Li}_2\left(-\infty\right)}{2}-\dfrac{\operatorname{Li}_2\left(1-\infty\right)}{2}\right]-\left[\dfrac{\ln\left(1\right)\ln\left(1+1\right)}{2}-\dfrac{\ln^2\left(1\right)}{2}+\dfrac{\operatorname{Li}_2\left(-1\right)}{2}-\dfrac{\operatorname{Li}_2\left(1-1\right)}{2}\right]$$
$$=\left[\underbrace{\dfrac{\operatorname{Li}_2\left(-\infty\right)}{2}-\dfrac{\operatorname{Li}_2\left(1-\infty\right)}{2}}_{=0}\right]-\left[\underbrace{\dfrac{\operatorname{Li}_2\left(-1\right)}{2}}_{-\pi^2/24}-\underbrace{\dfrac{\operatorname{Li}_2\left(0\right)}{2}}_{0}\right]$$
Finally we see that our integral is equal to $\frac{\pi^2}{24}.$
A: Another possibility, using the Basel problem formula $\sum_{k\geq 1}k^{-2}=\pi^2/6$.
We have
$$
I=-\int_0^{\pi/2}\frac{\cos(x)}{\sin(x)}\ln\cos(x)dx=
-\frac 12\int_0^{\pi/2}\frac{\cos(x)}{\sin(x)}\ln(1-\sin^2 x) dx
$$
$$
=\frac 12\sum_{k\geq 1}\frac 1k\int_0^{\pi/2}\cos(x)\sin^{2k-1}(x) dx
$$
The integral in the summation is evaluated as
$$\int_0^{\pi/2}\cos(x)\sin^{2k-1}(x) dx=\frac 1{2k}\int_0^{\pi/2}d(\sin^{2k}(x))dx=\frac 1{2k}$$
Hence
$$
I=\frac 14\sum_{k\geq 1}\frac 1{k^2}=\frac{\pi^2}{24}.
$$
A: I'm achieving your approach :
Using the Identity I mentioned :
$$\tan^2 x=\sec^2 x-1$$
And :
$$\ln(\sec x) =t\Longleftrightarrow \sec x=e^t$$
So $\tan^2 x$ in function of $t$ is :
$$\tan^2 x=e^{2t}-1$$
And your integral is :
$$\int_0^\infty \frac{t}{e^{2t}-1}dt=\frac{1}4\int_0^\infty \frac{t}{e^t-1}dt$$
Note that :
$$\zeta(s)\Gamma(s)=\int_0^\infty \frac{t^{s-1}}{e^t-1}dt$$
In our case :
\begin{align}
\zeta(2)\Gamma(2)&=\int_0^\infty  \frac{t}{e^t-1}dt\\
\frac{\pi^2}{6}\times 1&=\int_0^\infty  \frac{t}{e^t-1}dt\\
\end{align}
Therefore :
\begin{align}
I&=\frac{1}4 \int_0^\infty \frac{t}{e^t-1}dt\\
&=\frac{1}4\times \frac{\pi^2}{6}\\
&=\frac{\pi^2}{24}
\end{align}
Which is the result @DavidG.Strock said the answer Mathematica gives.

$\zeta(s)$ denotes Riemann Zeta function, $\Gamma(s)$ denotes Euler Gamma function.
$$\zeta(2)=\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$$
$$\Gamma(2)=(2-1)!=1$$
A: \begin{align} 
\int_0^{\pi/2} &\cot x\ln(\sec x) \, dx
=\frac14\int_0^{\pi/2}\frac{-2\cos x\sin x \ln(\cos^2 x)}{1-\cos^2 x} \, dx
\\
& \overset{\color{blue}{\overset{\scriptstyle t=\cos^2 x}{dt = -2\cos x\sin x\,dx}}}{=}\quad
\frac14\int_0^1\frac{\ln t}{t-1}\, dt
=\frac14\cdot\frac{\pi^2}6=\frac{\pi^2}{24}
.
\end{align}
