Answer
the curve:
$$ r=\cos^{4}( \frac{\theta }{4}) $$
is completely traced with
$$
\theta =0 \quad \text {to} \quad \theta=4 \pi
$$
the exact length of the given curve is equal to
\begin{aligned} L &=\int_{0}^{4\pi} \sqrt{r^{2}+(d r / d \theta)^{2}} d \theta \\
&=\frac{16}{3} \end{aligned}
Work Step by Step
The curve:
$$ r=\cos^{4}( \frac{\theta }{4}) $$
is completely traced with
$$
\theta =0 \quad \text {to} \quad \theta=4 \pi
$$
is equal to
$$
\begin{aligned} r^{2}+(d r / d \theta)^{2} &=\left[\cos ^{4}(\theta / 4)\right]^{2}+\left[4 \cos ^{3}(\theta / 4) \cdot(-\sin (\theta / 4)) \cdot \frac{1}{4}\right]^{2} \\ &=\cos ^{8}(\theta / 4)+\cos ^{6}(\theta / 4) \sin ^{2}(\theta / 4) \\ &=\cos ^{6}(\theta / 4)\left[\cos ^{2}(\theta / 4)+\sin ^{2}(\theta / 4)\right] \\
&=\cos ^{6}(\theta / 4) \end{aligned},
$$
then the exact length of the given curve is equal to
$$
\begin{aligned} L &=\int_{a}^{b} \sqrt{r^{2}+(d r / d \theta)^{2}} d \theta \\
& =\int_{0}^{4 \pi} \sqrt{\cos ^{6}(\theta / 4)} d \theta \\
&=\int_{0}^{4 \pi}\left|\cos ^{3}(\theta / 4)\right| d \theta \\ &=2 \int_{0}^{2 \pi} \cos ^{3}(\theta / 4) d \theta \\
& \quad\quad \quad\left[\operatorname{since} \cos ^{3}(\theta / 4) \geq 0 \text { for } 0 \leq \theta \leq 2 \pi\right] \\
& \quad\quad\quad\left[u=\frac{1}{4} \theta\right]\\
& =8 \int_{0}^{\pi / 2} \cos ^{3} u d u \\ &=8 \int_{0}^{\pi / 2}\left(1-\sin ^{2} u\right) \cos u d u \\
& \quad\quad\quad\left[\begin{array}{c}{x=\sin u} \\ {d x=\cos u d u}\end{array}\right] \\
&=8 \int_{0}^{1}\left(1-x^{2}\right) d x \\
&=8\left[x-\frac{1}{3} x^{3}\right]_{0}^{1} \\
& =8\left(1-\frac{1}{3}\right) \\
&=\frac{16}{3} . \end{aligned}
$$
