<s>[INST] integrate $\int \cos^{4}x\sin^{4}xdx$ 
$$\int \cos^4x\sin^4xdx$$

How should I approach this?
I know that $\sin^2x={1-\cos2x\over 2}$ and $\cos^2x={1+\cos2x\over 2}$
 [/INST] we now that
$$ \sin 2x=2\cos x\sin x$$
$$\cos x\sin x=\frac{\sin 2x}{2}$$
so...
$$\cos^4 x\sin^4 x=\frac{\sin^4 2x}{16}$$
$$\frac{\sin^4 2x}{16}=\frac{(\frac{1-\cos 4x}{2})^2}{16}=\frac{1-2\cos 4x+\cos^24x}{64}=\frac{1-2\cos 4x+\frac{1+\cos 8x}{2}}{64}$$
</s>