Answer
One loop of the curve
$$
r=\cos 2\theta
$$
is traced with
$$
-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}
$$
the length of one loop of the given curve is equal to
$$
\begin{aligned} L &=\int_{-\pi / 4}^{\pi / 4} \sqrt{r^{2}+(d r / d \theta)^{2}} d \theta \\
& \approx 2.4221
\end{aligned}
$$
Work Step by Step
One loop of the curve
$$
r=\cos 2\theta
$$
is traced with
$$
-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}
$$
$$
\begin{aligned}
r^{2}+\left(\frac{d r}{d \theta}\right)^{2}& =\cos ^{2} 2 \theta+(-2 \sin 2 \theta)^{2} \\
&=\cos ^{2} 2 \theta+4 \sin ^{2} 2 \theta \\
& =1+3 \sin ^{2} 2 \theta
\end{aligned}
$$
then the length of one loop of the given curve is equal to
$$
\begin{aligned} L &=\int_{a}^{b} \sqrt{r^{2}+(d r / d \theta)^{2}} d \theta \\
&=\int_{-\pi / 4}^{\pi / 4} \sqrt{1+3 \sin ^{2} 2 \theta} d \theta \\
& \approx 2.4221
\end{aligned}
$$
