Line-integral: 2-form integrated over cube 
Let $$\omega=x\cos(xy)\cos(2\pi x)\ \ \ \text{d}x\wedge\text{d}y$$ Calculate $$\int_{[0,\frac{1}{4}]\times[0,2\pi]}\omega$$

Now we have:
\begin{align} \int_{[0,\frac{1}{4}]\times[0,2\pi]}x\cos(xy)\cos(2\pi x)\  \text{d}x\wedge\text{d}y & = \int_0^{\frac{1}{4}}\Big(\int_0^{2\pi}x\cos(xy)\cos(2\pi x) \text{d}y\Big)\ \text{d}x\\
&= \int_0^{\frac{1}{4}}\cos(2\pi x)\sin(2\pi x)\\
&=\frac{1}{4}\pi
\end{align}
I am not sure if this is correct, could anyone have a look at it and spot any mistakes? I would appreciate it a lot!

It should be:
$$
\begin{align}
\int\limits_0^\frac{1}{4} \cos(2\pi x)\sin(2\pi x) \mathrm{d}x &= \int\limits_0^\frac{1}{4}\frac{1}{2}\sin(4\pi x)\mathrm{d}x \\
&= \frac{1}{2}\left[\frac{-1}{4\pi}\cos(4\pi x)\right]_0^\frac{1}{4} \\
&= \frac{1}{2}\cdot\frac{1}{4\pi}(1 - (-1)) \\
&= \frac{1}{4\pi}
\end{align}
$$.
