These solutions are all very well thought out, but an integration bee problem has to be solved in minutes. Most likely the intent behind the problem was to recognize that
$$\int_0^\pi \sin^2(3x+\cos^4(5x))\:dx = \int_0^\pi\sin^2(3x)\:dx$$
or in other words the cosine term doesn't matter, which solves the problem immediately. But how would one quickly justify that to themselves?
Consider
$$I[a] = \int_0^\pi\sin^2(3x+a\cos^4(5x))\:dx \implies I'[a] = \int_0^\pi\sin(6x+2a\cos^4(5x))\cos^4(5x)\:dx$$
There are a few immediate simplifications to be made here. First using the sine angle addition formula
$$\require{cancel} I'[a] = \cancelto{0}{\int_0^\pi\sin(6x)\cos(2a\cos^4(5x))\cos^4(5x)\:dx}+\int_0^\pi\cos(6x)\sin(2a\cos^4(5x))\cos^4(5x)\:dx$$
which cancels due to odd symmetry on the interval. Next, observe that the remaining integrand is $\pi$-periodic, so we can move the integral over any interval of $\pi$ length we wish. Choosing $[0,\pi]\to\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ we get
$$I'[a] = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(6x)\sin(2a\cos^4(5x))\cos^4(5x)\:dx = \cancelto{0}{2\int_0^{\frac{\pi}{2}}\cos(6x)\sin(2a\cos^4(5x))\cos^4(5x)\:dx}$$
due to even then odd symmetry. Thus $I[a]$ is a constant so
$$I[a] = I[0] = \int_0^\pi\sin^2(3x)\:dx = \frac{\pi}{2}$$
Explaining the steps out makes this seem like a long process, but on a bee contestants would not have to think the symmetry computations all the way through to justify to themselves to a good degree of confidence that things cancel out. The same cannot be said for computations involving series.