How would one prove that:
$$\int_0^{\pi/2} \frac{\ln (1+\cos \theta)}{\cos \theta}\, {\rm d}\theta= \frac{\pi^2}{8}$$
This is what I did.
\begin{align*} \int_{0}^{\pi/2}\frac{\ln \left ( 1+\cos \theta \right )}{\cos \theta}\, {\rm d}\theta &= \int_{0}^{\pi/2}\frac{1}{\cos \theta} \sum_{n=1}^{\infty}\frac{(-1)^{n-1} \cos^n \theta}{n} \, {\rm d}\theta \\ &= \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} \int_{0}^{\pi/2} \cos^{n-1}\theta \, {\rm d}\theta\\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{2n}{\rm B}\left ( \frac{1}{2}, \frac{n}{2} \right ) \end{align*}
No clue how to proceed. I am also interest in seeing another approach that does not use Taylor expansion.
I give some steps here, and I leave it to you to fill in the details.
Start with $$ f(a)=\int_0^{\pi/2}\frac{\ln(1+a \cos \theta)}{\cos\theta}\,d\theta. $$ Then $$ f'(a)=\int_0^{\pi/2}\frac{1}{1+a\cos \theta}\,d\theta=\cdots=\frac{2}{\sqrt{1-a^2}}\arctan\Bigl(\frac{\sqrt{1-a}}{\sqrt{1+a}}\Bigr) $$ Integrating, we find that $$ \begin{aligned} f(a)&=f(0)+\int_0^af'(\theta)\,d\theta=0+\Bigl[-2\Bigl\{\arctan\Bigl(\frac{\sqrt{1-\theta}}{\sqrt{1+\theta}}\Bigr)\Bigr\}^2\Bigr]_0^a\\ &=\frac{\pi^2}{8}-2\Bigl\{\arctan\Bigl(\frac{\sqrt{1-a}}{\sqrt{1+a}}\Bigr)\Bigr\}^2. \end{aligned} $$ In particular $$ \int_0^{\pi/2}\frac{\ln(1+ \cos \theta)}{\cos\theta}\,d\theta=f(1)=\frac{\pi^2}{8}. $$