Evaluate $\int_{0}^{\frac{\pi}{4}} \ln(\sec x)dx$ Evaluate $$P=\int_{0}^{\frac{\pi}{4}} \ln(\sec x)dx$$
My try: I tried using its complimentary integral:
Let $$Q=\int_{0}^{\frac{\pi}{4}} \ln(\csc x)dx$$
Adding both we get:
$$P+Q=\int_{0}^{\frac{\pi}{4}}\ln(\sec x\csc x)dx$$ $\implies$
$$2P+2Q=\int_{0}^{\frac{\pi}{4}}\ln(\sec^2 x\csc^2 x)dx=\int_{0}^{\frac{\pi}{4}}\ln\left(\frac{4}{4\sin^2 x\cos^2 x}\right)dx$$ $\implies$
$$2P+2Q=\frac{\pi}{4}\ln 4-\int_{0}^{\frac{\pi}{4}}\ln\left(\sin^2 2x\right)dx$$
$$2P+2Q=\frac{\pi}{2}\ln 2-2 \int_{0}^{\frac{\pi}{4}}\ln(\sin 2x)dx$$
Using the substitution $2x=t$ we get
$$2P+2Q=\frac{\pi}{2}\ln 2- \int_{0}^{\frac{\pi}{2}}\ln(\sin t)dt$$
Using the formula:
$$\int_{0}^{\frac{\pi}{2}}\ln(\sin t)dt=\frac{-\pi}{2}\ln 2$$ we get
$$2P+2Q=\pi \ln 2$$
$$P+Q=\frac{\pi}{2}\ln 2$$
Is there any way to find $P-Q$

Through the properties of the logarithm, the integral can be rewritten as$$\mathfrak{I}=-\int\limits_0^{\pi/4}\mathrm dx\,\log\cos x$$Now use the Fourier expansion for $\log\cos x$ which is

$$\log\cos x=\sum\limits_{n\geq1}\frac {(-1)^{n-1}\cos 2nx}{n}-\log 2$$

Now integrate each time termwise to get that$$\begin{align*}\mathfrak{I} & =\frac {\pi}4\log 2+\sum\limits_{n\geq1}\frac {(-1)^n}{n}\int\limits_0^{\pi/4}\mathrm dx\,\cos 2nx\\ & =\frac {\pi}4\log 2+\frac 12\sum\limits_{n\geq1}\frac {(-1)^n}{n^2}\sin\left(\frac {\pi n}2\right)\\ & =\frac {\pi}4\log 2-\frac 12G\end{align*}$$where $G$ is Catalan’s constant.
