Solving: $\int\cos^4(x)\sin^4(x)\ dx$ I've been trying to solve this problem - $\int\cos^4(x)\sin^4(x)\ dx$, but I don't seem to be succeeding. This is what I've done: $$\int\left(\frac{1+\cos(2x)}{2}\right)^2\left(\frac{1-\cos(2x)}{2}\right)^2\ dx\\=\frac1{16}\int\left(1+2\cos(2x)+\cos^2(2x)\right)\left(1-2\cos(2x)+\cos^2(2x)\right)\ dx\\=\frac1{16}\int1+2\cos^2(2x)+\cos^4(2x)\ dx\\=\frac1{16}\int1+1+\cos(4x)+\frac{
(1+\cos(4x))^2}{4}\ dx\\=\frac1{16}\int2+\cos(4x)+\frac{1+2\cos(4x)+\cos^2(4x)}{4}\ dx\\=\frac1{16}\int2+\cos(4x)+\frac{1}{4}+\frac{\cos(2x)}{2}+\frac{1+\cos(8x)}{8}\ dx\\=\frac1{16}\left(2x+\frac{\sin(4x)}{4}+\frac{x}4+\frac{\sin(2x)}{4}+\frac{x}8+\frac{\sin(8x)}{64}\right)\\=\frac{x}{8}+\frac{\sin(4x)}{64}+\frac{x}{64}+\frac{\sin(2x)}{64}+\frac{x}{128}+\frac{\sin(8x)}{1024}\\=\frac{19x}{128}+\frac{\sin(2x)}{64}+\frac{\sin(4x)}{64}+\frac{\sin(8x)}{1024}$$
I checked this up on integral-calculator.com. They seem to have a similiar answer - but not quite the same - their denominators are different and they don't have a $\sin(2x)$ term. I keep retrying this problem, but I seem to be missing something. Where am I going wrong?
 A: Generally 
$$
(1+2a+b)(1-2a+b)\neq 1+2a^2+b^2,
$$
as it is in your calculations (lines 2 and 3) with $a=\cos(2x)$ and $b=\cos^2(2x)$.
Probably a quicker way is to use $\sin2x=2\sin x\cos x$ and reduce to solving integral of the type
$$
\int\sin^4x\,dx.
$$
A: It is much simpler (and shorter) to linearise through the complex exponential:
First rewrite this function as
$$(\cos x\sin x)^4=\frac 1{2^4}\sin^4 2x.$$
Next linearise:
\begin{align}
\sin^4 2x&=\frac1{(2i)^4}\bigl(\mathrm e^{2ix}-\mathrm e^{-2ix}\bigr)^4=\frac1{16}\bigl(\mathrm e^{8ix}-4\mathrm e^{4ix}+6-\mathrm e^{-4ix}+\mathrm e^{-8ix}\bigr)\\
&=\frac1{16}(2\cos 8x-8\cos 4x+6)=\frac18(\cos 8x-4\cos 4x+3).
\end{align}
A: Hint: An alternative approach is to use the identity 
$$\sin(2x) = 2\sin(x)\cos(x)$$
Then we get that
\begin{align*}
\int \cos^4(x)\sin^4(x)dx &= \int \big(\cos(x)\sin(x)\big)^4 dx\\
&= \int \left(\frac{\sin(2x)}{2}\right)^4 dx \\
&= \frac{1}{16} \int \sin^2(2x) \sin^2(2x)dx \\
&= \frac{1}{16} \int \left(\frac{1-\cos(4x)}{2}\right)\left(\frac{1-\cos(4x)}{2}\right)dx
\end{align*}
And then the rest I leave as an exercise. 
