Example 8. Find the length of the arc of the curve $r=a \cos ^{3} \frac{\varphi}{3}(a>0)$.
Solution. Since $r \geqslant 0$, then $\cos ^{3} \frac{\varphi}{3} \geqslant 0$, which means $\frac{\varphi}{3} \in \left[-\frac{\pi}{2} ; \frac{\pi}{2}\right]$, or $\varphi \in \left[-\frac{3 \pi}{2} ; \frac{3 \pi}{2}\right]$. The cosine function is an even function, so we will limit ourselves to calculating the length corresponding to the interval $\varphi \in \left[0 ; \frac{3 \pi}{2}\right]$, and double the result.

We will use formula (5): $L=2 \int_{0}^{3 \pi / 2} \sqrt{r^{2}+\left(r^{\prime}\right)^{2}} d \varphi$. We have:

$$
r^{\prime}(\varphi)=-a \cos ^{2} \frac{\varphi}{3} \cdot \sin \frac{\varphi}{3}, \sqrt{r^{2}+\left(r^{\prime}\right)^{2}}=a \cos ^{2} \frac{\varphi}{3}
$$

and

$$
\begin{aligned}
L=2 a \int_{0}^{3 \pi / 2} \cos ^{2} \frac{\varphi}{3} d \varphi=a \int_{0}^{3 \pi / 2}\left(1+\cos \frac{2 \varphi}{3}\right) & d \varphi= \\
& =\left.a\left(\varphi+\frac{3}{2} \sin \frac{2 \varphi}{3}\right)\right|_{0} ^{3 \pi / 2}=\frac{3 \pi}{2} a
\end{aligned}
$$

Answer. $L=\frac{3 \pi}{2} a$.