## Calculus: Early Transcendentals 8th Edition

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}
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}