Answer
$$\int_{0}^{\pi}\sqrt{\sin^2\theta\cos^2\theta+\cos^22\theta} d \theta\\
=\int_{0}^{\pi}\sqrt{\frac{1}{4}\sin^22\theta+\cos^22\theta} d\theta $$
Work Step by Step
Since $r=f(\theta)=\sin\theta\cos\theta$, then $f'(\theta)=\cos^2\theta-\sin^2\theta=\cos2\theta$
Thus, the length is given by
$$ \text { Length }=\int_{0}^{\pi} \sqrt{f(\theta)^{2}+f^{\prime}(\theta)^{2}} d \theta\\ =\int_{0}^{\pi}\sqrt{\sin^2\theta\cos^2\theta+\cos^22\theta} d \theta\\
=\int_{0}^{\pi}\sqrt{\frac{1}{4}\sin^22\theta+\cos^22\theta} d\theta $$
