Trig Integral (Definite) How do I evaluate this definite integral? 
$$\int_{0}^{\frac{\pi}{12}}{\sin^4x \, \cos^4x\, \operatorname{d}\!x}$$
I know this is a trig. function. 
 A: From here, we have
$$\int_0^{\pi/2} \sin^{2n}(x) dx = \dfrac{2n-1}{2n} \cdot \dfrac{2n-3}{2n-2} \cdots \dfrac34 \cdot \dfrac12 \cdot\dfrac{\pi}2$$
Hence,
$$\int_0^{\pi/2} \sin^4(x) \cos^4(x) dx = \dfrac1{16} \int_0^{\pi/2} \sin^4(2x) dx = \dfrac1{32} \int_0^{\pi} \sin^4(t) dt = \dfrac2{32} \int_0^{\pi/2} \sin^4(t) dt$$
Hence, the answer is
$$\dfrac1{16} \cdot \dfrac34 \cdot \dfrac12 \cdot \dfrac{\pi}2 = \dfrac{3 \pi}{256}$$

If the integral is from $0$ to $\pi/12$, then, we get that
\begin{align}
\int_0^{\pi/{12}} \sin^4(x) \cos^4(x) dx & = \dfrac1{16} \int_0^{\pi/12} \sin^4(2x) dx = \dfrac1{32} \int_0^{\pi/6} \sin^4(t) dt\\
& = \dfrac1{32} \int_0^{\pi/6} \dfrac{(1-\cos(2t))^2}{4} dt\\
& = \dfrac1{128} \int_0^{\pi/6} (1+\cos^2(2t) - 2 \cos(2t)) dt\\
& = \dfrac1{128} \int_0^{\pi/6} \left(1+\dfrac{1+\cos(4t)}2 - 2 \cos(2t) \right) dt\\
& = \dfrac1{128} \int_0^{\pi/6} \left(\dfrac32+\dfrac{\cos(4t)}2 - 2 \cos(2t) \right) dt\\
& = \dfrac1{128} \left(\dfrac{\pi}4 + \dfrac{\sin(4 \pi/6)}{8}-\sin(2 \pi/6)\right)\\
& = \dfrac1{128} \left(\dfrac{\pi}4 + \dfrac{\sqrt3}{16} - \dfrac{\sqrt3}2\right)\\
& = \dfrac1{128} \left(\dfrac{\pi}4 - \dfrac{7\sqrt3}{16} \right) = \dfrac{4 \pi - 7 \sqrt3}{2048}
\end{align}
A: We have 
$$\sin^4x\cos^4x=\frac{1}{16}\sin^4(2x)=\frac{1}{16}(\frac{e^{2ix}-e^{-2ix}}{2i})^4=\frac{1}{16^2}(2\cos(8x)-8\cos(4x)+6), $$
so we integrate and we find
$$\int_{0}^{\frac{\pi}{12}}{\sin^4x \, \cos^4x\, \operatorname{d}\!x}=\frac{1}{512}π-\frac{7}{2048}\sqrt{3}.$$
