What does $\int \log ( \operatorname{sech}(\log x))dx$ have to do with the area enclosed by $r = \sin(2 \theta)$ and $ r=\cos(2 \theta)$? Let $f(x) := -\log(\operatorname{sech}(\log(x)))$. Also, let $A$ be the area enclosed by the curves $r=\sin(2\theta)$ and $r=\cos(2\theta)$, overlap not included twice. I have shown that
$$\int_0^1f(x)~dx= \frac \pi 2-1 =A.$$
I am wondering if this can be understood intuitively, if this is a coincidence, or if there is some generalization of this. I don't quite understand what the area covered by two relatively simple polar curves has to do with a function like $f(x)$, but nonetheless am interested in whether there is a larger result making an appearance here.
 A: 
There does not seem to be any natural or intuitive connections between the integral and the area. 
As can be seen from the graph, the enclosed area is of a 8-paddle flower. The first paddle tip point P is the intersection of the two curves at the polar angle $\frac\pi8$. Then, the enclosed area can be integrated directly as $$8\int_0^{\pi/8}\int_0^{\sin2\theta}2rdr d\theta=8\int_0^{\pi/8}\sin^2(2\theta)d\theta=\frac\pi2 -1$$
On the other hand, the given integral can be evaluated with integration by parts,
$$I = -\int_0^1 \ln \operatorname{sech}\ln x~dx = - x\ln \operatorname{sech}\ln x|_0^1 - \int_0^1 \tanh \ln x dx = \int_0^\infty e^{-t}\tanh t dt$$
where the substitution $\ln x = -t$ was made in the last step. Then,
$$I= \int_0^\infty \frac{e^t-e^{-t}}{e^t+e^{-t}} e^{-t}dt
= \int_0^\infty \left( -e^{-t} + \frac{2e^{-t}}{1+e^{-2t}} \right)dt
=(e^{-t}-2\tan^{-1}e^{-t})|_0^\infty=\frac\pi2 -1$$
which happens to have the same analytic value as that of the enclosed area and appears to be just coincidental. 
