Integrate $\frac{\sin^2(2t)}{4}$ from $0$ to $2\pi$ Im trying to integrate $$\int_{0}^{2\pi}\dfrac{\sin^2(2t)}{4} dt$$  and I'm stuck.. 
Im at the end of multivariable calculus so no good solutions for these "simpler" problems are shown in the book at this point. 

you can also use the fact that $$ \int_0^{2\pi} \sin^2(2x) \, dx = \int_0^{2\pi}\cos^2 (2x)\, dx \text{ and } \int_0^{2\pi} (\sin^2(2x) + \cos^2(2x))\, dx = \int_0^{2\pi} 1 \, dx = 2\pi$$ to conclude $$\int_0^{2\pi} \frac{\sin^2(2x)}{4} \, dx = \frac{\pi}4.$$
