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. 

$$\sin^2(\alpha)=\frac12-\frac12\cos2\alpha$$ oscillates symmetrically around $\dfrac12$. Integrating over an integer number of periods, we have $\dfrac14\dfrac122\pi$.
