<s>[INST] Arc Length polar curve $$r=a\sin^3\left(\frac{\theta}{3}\right) $$
I tried solving it using the equation for arc length with $dr/d\theta$ and $r^2$. Comes out messy and complicated. 
 [/INST] The portion of the arclength calculation in which we determine the infinitesimal arclength element $ \ ds \ $ just requires applying the Chain Rule carefully and using the Pythagorean Identity:
$$ \frac{dr}{d\theta} \ = \ \frac{d}{d\theta} \left[ \ a \ \sin^3 \left(\frac{\theta}{3}\right) \right] \ = \ a \ \cdot \ 3 \ \sin^2\left(\frac{\theta}{3}\right)  \cdot  \cos\left(\frac{\theta}{3}\right)  \cdot  \frac{1}{3} \ = \ a \  \sin^2\left(\frac{\theta}{3}\right)  \cos\left(\frac{\theta}{3}\right)  $$
$$ \Rightarrow \ \ ds \ = \ \sqrt{ \ r^2  +  \left(\frac{dr}{d\theta}\right)^2} \ \ d\theta \ = \ \sqrt{ \ \left[ \ a \ \sin^3 \left(\frac{\theta}{3}\right) \right]^2  +  \left[ \ a \  \sin^2\left(\frac{\theta}{3}\right)  \cos\left(\frac{\theta}{3}\right) \right]^2} \ \ d\theta  $$
$$  = \ \sqrt{   \ a^2 \ \sin^6 \left(\frac{\theta}{3}\right)   +  \ \ a^2 \  \sin^4\left(\frac{\theta}{3}\right)  \cos^2\left(\frac{\theta}{3}\right)} \ \ d\theta $$
$$  = \ \sqrt{ \ a^2 \ \sin^4 \left(\frac{\theta}{3}\right) \left[ \ \sin^2 \left(\frac{\theta}{3}\right)   +  \   \cos^2\left(\frac{\theta}{3}\right) \right]} \ \ d\theta $$
$$ = \ \sqrt{ \ a^2 \ \sin^4 \left(\frac{\theta}{3}\right) } \ \ d\theta \ = \ a \ \sin^2 \left(\frac{\theta}{3}\right)  \ d\theta \ \ . $$
What turns out to be the tricky part is in finding the limits of integration in order to cover this curve fully.  Here is a graph of the curve described by this polar equation (for $ \ a \ = \ 1 ) \ $ :

A graph of $ \ r \ $ as a function of $ \ \theta \ $ shows that, while the period of the function is $ \ 6 \pi \ $ , the radius is negative in the interval $ \ ( \ 3 \pi \ , \ 6 \ \pi \ ) \ $ .

The curve resembles a limaçon , but the interval of negative radii retraces the curve already "swept out" by positive radii in $ \ ( \ 0 \ , \ 3 \pi \ ) \ $ .  Thus, we will find the arclength correctly by integrating arclength over this interval, rather than only up to $ \ 2 \pi \ $ or all the way to $ \ 6 \pi \ $ .
Our arclength integral is therefore
$$ \int_0^{3 \pi}  \ ds \ \ = \ \ \int_0^{3 \pi}  a \ \sin^2 \left(\frac{\theta}{3}\right) \ \ d\theta \ \ = \ \ a \ \int_0^{3 \pi} \ \frac{1}{2} \left[ \ 1 \ - \   \cos \left(\frac{2\theta}{3}\right)  \right] \ \ d\theta  $$
$$ = \ \  \frac{a}{2}   \left[ \ \theta \ - \   \frac{3}{2} \sin \left(\frac{2\theta}{3}\right)  \right] \vert_0^{3 \pi} \ = \  \frac{a}{2}   \left[ \ \left(3 \pi \ - \ \frac{3}{2} \sin \left[\frac{6 \pi}{3}\right] \ \right) \ - \ \left(0 \ - \ \frac{3}{2} \sin \left[\frac{0}{3}\right] \right)  \right] $$
$$ = \  \frac{a}{2}    \ \left(3 \pi \ - \ \frac{3}{2} \sin \ 2 \pi \  \ - \ 0 \ + \ 0 \right) \ = \ \frac{3 \pi}{2} a \ \ .  $$
As a check, we can see from the graph above that the arclength of the curve will be somewhat larger than the circumference of a circle of diameter $ \ \frac{5}{4} a \ $ , which is $ \ \frac{5 \pi}{4} a \ $ (the small "self-crossing" loop of this curve does not add much to the total arclength).
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