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. 

$$\int_{0}^{2\pi}\frac{1}{4}sin^2(2t)dt = \frac{2}{4}\int_{0}^{\pi}sin^2(2t) = \frac{t-\sin(t)\cos(t)}{4}|_{0}^{\pi} = \pi/4$$
