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 $\cos^2x={1+\cos2x\over 2}$
 A: Use
$$
\cos x=\frac{e^{ix}+e^{-ix}}{2},
\qquad
\sin x=\frac{e^{ix}-e^{-ix}}{2i}
$$
so
\begin{align}
\cos^4x\sin^4x
&=\frac{1}{2^8}(e^{2ix}-e^{-2ix})^4\\[6px]
&=\frac{1}{2^8}(e^{8ix}-4e^{4ix}+6-4e^{-4ix}+e^{8ix})\\[6px]
&=\frac{1}{128}\cos8x-\frac{1}{32}\cos4x+\frac{3}{128}
\end{align}
A: Continuing from where you left off:
$$\sin^2x\cos^2x=\left({1-\cos2x\over 2}\right)\left({1+\cos2x\over 2}\right)$$
$$\sin^2x\cos^2x=\frac{1}{4}(1 - \cos^2(2x))$$
$$\sin^2x\cos^2x=\frac{1}{4}(\sin^2(2x))$$
Squaring both the sides:
$$\sin^4x\cos^4x=\frac{1}{16}(\sin^4(2x))$$
This can be integrated in two ways:
Method 1:
$$\sin^4x\cos^4x=\frac{1}{16}(\sin^4(2x))$$
$$=\frac{1}{16}(\sin^4(2x)) = \frac{1}{16} \sin^2{(2x)} (1 - \cos^2{(2x)})$$
$$=\frac{1}{16} \left[\sin^2{(2x)} - \sin^2{(2x)} \cos^2{(2x)}\right]$$
$$=\frac{1}{16} \left[\sin^2{(2x)} - \frac{1}{4}(1 - \cos^2{(4x)})\right]$$
$$=\frac{1}{16} \left[\sin^2{(2x)} - \frac{1}{4}\sin^2{(4x)}\right] $$
$$=\frac{1}{32}[1 - \cos{(4x)}] - \frac{1}{128}[1 - \cos{(8x)}]$$
Which is easy to integrate
Method 2:
$$\:\frac{1}{16}\left[\sin^2(2x)\right]^2\;\;$$
$$=\frac{1}{16}\left[\frac{1\,-\,\cos(4x)}{2}\right]^2\:$$
$$=\:\frac{1}{64}\left[1\,-\,2\cdot\cos(4x) \,+\,\cos^2(4x)\right]$$
$$=\:\frac{1}{64}\left[1\,-\,2\cdot\cos(4x) \,+\,\frac{1+\cos(8x)}{2}\right]$$
$$=\:\frac{1}{128}\left[3\,-\,4\cdot\cos(4x)\,+\,\cos(8x)\right]$$
Which is again easy to integrate
A: we now that
$$ \sin 2x=2\cos x\sin x$$
$$\cos x\sin x=\frac{\sin 2x}{2}$$
so...
$$\cos^4 x\sin^4 x=\frac{\sin^4 2x}{16}$$
$$\frac{\sin^4 2x}{16}=\frac{(\frac{1-\cos 4x}{2})^2}{16}=\frac{1-2\cos 4x+\cos^24x}{64}=\frac{1-2\cos 4x+\frac{1+\cos 8x}{2}}{64}$$
