I am trying to obtain a recurrence relation for the following integral, $$I_n=\int_0^{\pi/2}\log^n(\sec^2 x)\ dx$$ using exponential generating functions.
The general structure of the integral seems to be suited for this method, but I ended up with a term which I could not get the Maclaurin coefficient of. I feel like this problem could be avoided if I took the logarithmic derivative at a different stage of the evaluation, but I think finding a series expansion of that would also be difficult.
Sorry for the long question; my attempt is below.
Define the exponential generating function, $$G(t)=\sum_{n=0}^\infty\frac{I_n}{n!}t^n=\int_0^{\pi/2}\sum_{n=0}^\infty\frac{(t\log(\sec^2x))^n}{n!}\ dx=\int_0^{\pi/2}\sec^{2t}(x)\ dx$$ by the integral representation of the Beta function, we can re write $G(t)$ in terms of the Gamma function, $$\int_0^{\pi/2}\cos^{-2t} (x)\ dx=\frac{1}{2}B\left(\dfrac{1}{2},\dfrac{1-2t}{2}\right)=\frac{\sqrt\pi}{2}\dfrac{\Gamma\left(\dfrac{1-2t}{2}\right)}{\Gamma\left(1-t\right)}$$ applying Legendre's duplication formula to the numerator, $$G(t)=2^{2t-1}\pi\frac{\Gamma(1-2t)}{\Gamma^2(1-t)}$$ by Euler's reflexion formula, $$\Gamma(1-2t)=\frac{\pi}{\sin(2\pi t)\Gamma(2t)},\quad\Gamma(1-t)=\frac{\pi}{\sin(\pi t)\Gamma(t)}$$ we find, $$G(t)=2^{t-2}\tan(\pi t)\frac{\Gamma^2(t)}{\Gamma(2t)}=\frac{\sqrt\pi}{2}\frac{\Gamma(t)\tan(\pi t)}{\Gamma\left(t+\dfrac{1}{2}\right)}$$ where we used Legendre's duplication formula again. Taking the logarithmic derivative, $$\begin{align*}\frac{G'(t)}{G(t)}&=\psi(t)-\psi\left(t+\frac{1}{2}\right)+\frac{\pi}{\sin (\pi t)\cos (\pi t)} \\ &=\psi(t)-\psi\left(t+\frac{1}{2}\right)+2\pi \color{red}{\csc(2\pi t)}\end{align*}$$ where is the expansion, $$\psi(t)-\psi\left(t+\frac{1}{2}\right)=\sum_{n=1}^\infty (-1)^{n+1}\left(1-\frac{1}{2^n}\right)\zeta(n+1)t^n$$ but I can't get a Maclaurin series for $\color{red}{\csc(2\pi t)}$. If the red term somehow had an expansion, the recurrence relation is obtained after multiplying by $G(t)$ on both sides and using the property of EGF: $$[t^n/n!]G'(t)=I_{n+1}.$$
$$\csc (x)=\sum_{n=0}^\infty (-1)^{n+1}\,\frac{\left(4^n-2\right) B_{2 n}}{(2 n)!}\,x^{2n-1}$$