After multiple steps into a problem, I arrived at the integral
$$\int_0^{2\pi}\frac{100}{16+9\sin^2t}\mathrm{d}t.$$
Here the primitive can be easily calculated by posing $x=\arctan u$ and $\mathrm{d}x=\frac{1}{u^2+1}\mathrm{d}u$. We obtain that
$$\int\frac{100}{16+9\sin^2t}\mathrm{d}t=5\arctan\frac{5}{4}\tan t.$$
If the integration domain is applied to the primitive, I get
$$\int_0^{2\pi}\frac{100}{16+9\sin^2t}\mathrm{d}t=5\arctan\frac{5}{4}\tan 2\pi-5\arctan\frac{5}{4}\tan 0=0.$$
By looking at the graph of $\frac{100}{16+9\sin^2t}$, clearly the area under the curve in $[0,2\pi]$ isn't zero. This function is always positive and never crosses the x-axis. If I plug the initial integral directly into Maple, it gives me that
$$\int_0^{2\pi}\frac{100}{16+9\sin^2t}\mathrm{d}t=10\pi,$$
which is the expected answer in my problem. Now, I'm not sure what's wrong with the integral. As far as I can understand it, the integral is not Riemann integrable (correct me if I'm wrong), thus
$$\int_0^{2\pi}\frac{100}{16+9\sin^2t}\mathrm{d}t\neq 5\arctan\frac{5}{4}\tan 2\pi-5\arctan\frac{5}{4}\tan 0.$$
My question is the following, how does Maple compute this integral? Or in a more general sense, what to do when the integral isn't Riemann integrable? Which part of calculus did I break?