Double Integral Polar Coordinates in terms of r(θ) Bit of trouble with the following problem:

$\int\int_R 1 \,\mathrm{d}A$, where $0 \le r \le r(\theta)$. $r(\theta) = 8 − \sin(\theta) + 2 \sin(3\theta) + 2 \sin(5\theta)−\sin(7\theta) + 3 \cos(2\theta)−2 \cos(4\theta).$

So, I set up my integrals as: 
$\int_0^{2π}\int_0^{r(\theta)} r\,\mathrm{d}r\,\mathrm{d}\theta$. 
First, integrating with respect to $r$, you get $\frac{1}{2}(r(\theta))^2$, but when you integrate this with respect to $\theta$, it becomes a tricky integral.
Help please?
 A: $$\int_{0}^{2\pi}\int_{0}^{r\left(\theta\right)}r\,\mathrm{d}r\,\mathrm{d}\theta=\int_{0}^{2\pi}\dfrac{1}{2}\left(r(\theta)\right)^{2}\,\mathrm{d}\theta$$
$$=\frac{1}{2}\int_{0}^{2\pi}\left(8-\sin\theta+2\sin3\theta+2\sin5\theta-\sin7\theta+3\cos2\theta-2\cos4\theta\right)^{2}\,\mathrm{d}\theta$$
Now, this expands to many terms, but all of the cross terms turn out to have integral zero.
For $8$ and a single trig function, we have things like $\int_{0}^{2\pi}2*8\left(-\sin7\theta\right)\,\mathrm{d}\theta=0$ since we go through $7$ periods of sine and everything cancels out.
For products of two different trig functions, we can use the product to sum formulas  to reduce everything to sums of individual trig functions. For example, $\int_{0}^{2\pi}2*\left(2\sin5\theta\right)\left(3\cos2\theta\right)\,\mathrm{d}\theta=12\int_{0}^{2\pi}\frac{1}{2}\left(\sin7\theta+\sin3\theta\right)\,\mathrm{d}\theta$ and each part has integral zero.
The only terms that remain are the squares of the terms of $r(\theta)$. But, even then, things simplify. For example, $\sin^{2}3\theta=\frac{1}{2}-\frac{1}{2}\cos6\theta$ but the $\cos6\theta$ part integrates to $0$. Every square of a trig function just becomes $\frac12$ in that way because of the formulas for $\cos 2x$.
Therefore, the original integral reduces to:
$$\frac{1}{2}\int_{0}^{2\pi}\left(8^{2}+\frac{\left(-1\right)^{2}}{2}+\frac{2^{2}}{2}+\frac{2^{2}}{2}+\frac{\left(-1\right)^{2}}{2}+\frac{3^{2}}{2}+\frac{\left(-2\right)^{2}}{2}\right)\,\mathrm{d}\theta$$
$$=\frac{1}{2}\int_{0}^{2\pi}\left(\frac{128+1+4+4+1+9+4}{2}\right)\,\mathrm{d}\theta=\frac{1}{2}\int_{0}^{2\pi}\left(\frac{151}{2}\right)\,\mathrm{d}\theta=\frac{151}{2}\pi\text{.}$$
