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. 
 A: Hint: $$\sin^2 x = \frac{1}{2}\Big(1 - \cos 2x \Big)$$
Take $x = 2t$.
A: Hint: Note that by shifting by $\pi/4$ on the circle we get
$$
\begin{align}
\int_0^{2\pi}\frac{\sin^2(2t)}4\mathrm{d}t
&=\int_0^{2\pi}\frac{\cos^2(2t+\pi/2)}4\mathrm{d}t\\
&=\int_0^{2\pi}\frac{\cos^2(2t)}4\mathrm{d}t
\end{align}
$$
Now, consider the sum
$$
\int_0^{2\pi}\frac{\sin^2(2t)}4\mathrm{d}t+\int_0^{2\pi}\frac{\cos^2(2t)}4\mathrm{d}t
$$
A: Double angle formula: $cos(2A)=1-2{sin}^2(A)$
Rearrange to give an expression for ${sin}^2(A)$ in terms of $cos(2A)$
In your case $A=2t$
The result should be fairly easy to integrate.
A: $$\int_{0}^{2\pi}\frac{1}{4}sin^2(2t)dt = \frac{2}{4}\int_{0}^{\pi}sin^2(2t) = \frac{t-\sin(t)\cos(t)}{4}|_{0}^{\pi} = \pi/4$$
A: 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 $$\int_0^{2\pi} \frac{\sin^2(2x)}{4} \, dx = \frac{\pi}4.$$
A: $$\sin^2(\alpha)=\frac12-\frac12\cos2\alpha$$ oscillates symmetrically around $\dfrac12$. Integrating over an integer number of periods, we have $\dfrac14\dfrac122\pi$.
