Answer
The exact length of the polar curve:
$$ r=2(1+\cos \theta) \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$
is equal to
$$
\begin{aligned} L &=\int_{a}^{b} \sqrt{r^{2}+(d r / d \theta)^{2}} d \theta \\
& =\int_{0}^{2 \pi} \sqrt{[2(1+\cos \theta)]^{2}+(-2 \sin \theta)^{2}} d \theta\\
&=16 \end{aligned}
$$
Work Step by Step
The exact length of the polar curve:
$$ r=2(1+\cos \theta) \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$
is equal to
$$
\begin{aligned} L &=\int_{a}^{b} \sqrt{r^{2}+(d r / d \theta)^{2}} d \theta \\
& =\int_{0}^{2 \pi} \sqrt{[2(1+\cos \theta)]^{2}+(-2 \sin \theta)^{2}} d \theta \\ & =\int_{0}^{2 \pi} \sqrt{4+8 \cos \theta+4 \cos ^{2} \theta+4 \sin ^{2} \theta} d \theta \\
&=\int_{0}^{2 \pi} \sqrt{8+8 \cos \theta} d \theta \\
&=\sqrt{8} \int_{0}^{2 \pi} \sqrt{1+\cos \theta} d \theta \\
& =\sqrt{8} \int_{0}^{2 \pi} \sqrt{2 \cdot \frac{1}{2}(1+\cos \theta)} d \theta \\
&=\sqrt{8} \int_{0}^{2 \pi} \sqrt{2 \cos ^{2} \frac{\theta}{2}} d \theta \\
& =\sqrt{8} \sqrt{2} \int_{0}^{2 \pi}\left|\cos \frac{\theta}{2}\right| d \theta \\&
= 4 \cdot 2 \int_{0}^{\pi} \cos \frac{\theta}{2} d \theta \quad \text { [by symmetry] } \\
=& 8\left[2 \sin \frac{\theta}{2}\right]_{0}^{\pi} \\
& = 8(2)\\
&=16 \end{aligned}
$$
