"Integral milking": working backward to construct nontrivial integrals

I begin this post with a plea: please don't be too harsh with this post for being off topic or vague. It's a question about something I find myself doing as a mathematician, and wonder whether others do it as well. It is a soft question about recreational mathematics - in reality, I'm shooting for more of a conversation. I know that a lot of users on this site (e.g. Cleo, Jack D'Aurizio, and so on) are really good at figuring out crafty ways of solving recreational definite integrals, like $$\int_{\pi/2}^{\pi} \frac{x\sin(x)}{5-4\cos(x)} \, dx$$ or $$\int_0^\infty \bigg(\frac{x-1}{\ln^2(x)}-\frac{1}{\ln(x)}\bigg)\frac{dx}{x^2+1}$$ When questions like this pop up on MSE, the OP provides an integral to evaluate, and the answerers can evaluate it using awesome tricks including (but certainly not limited to): Clever substitution Exploitation of symmetry in the integrand Integration by parts Expanding the integrand as a series Differentiating a well-know integral-defined function, like the Gamma or Beta functions Taking Laplace and Inverse Laplace transforms But when I play around with integrals on my own, I don't always have a particular problem to work on. Instead, I start with a known integral, like $$\int_0^\pi \cos(mx)\cos(nx) \, dx=\frac{\pi}{2}\delta_{mn},\space\space \forall m,n\in \mathbb Z^+$$ and "milk" it, for lack of a better word, to see how many other obscure, rare, or aesthetically pleasing integrals I can derive from it using some of the above techniques. For example, using the above integral, one might divide both sides by $m$, getting $$\int_0^\pi \frac{\cos(mx)}{m}\cos(nx) \, dx=\frac{\pi}{2m}\delta_{mn},\space\space \forall m,n\in \mathbb Z^+$$ Then, summing both sides from $m=1$ to $\infty$, and exploiting a well-known Fourier Series, obtain $$\int_0^\pi \cos(nx)\ln(2-2\cos(x)) \, dx=-\frac{\pi}{n},\space\space \forall n\in \mathbb Z^+$$ or, after a bit of algebra, the aesthetically pleasing result $$\int_0^{\pi/2} \cos(2nx)\ln(\sin(x)) \, dx=-\frac