<s>[INST] Letting $r = 8\sin^2 (\theta/2)$ for $0 \leq\theta\leq\pi$. How do you find the length $L$ of the graph of the polar equation? Hello am having difficulty solving can someone please explain how this would be solved.
Thank you
 [/INST] $r = 8\sin^2\frac {\theta}{2} = 4 - 4\cos\theta$
And that is a cartioid.
$\int_0^{2\pi} (r^2 + (\frac {dr}{d\theta})^2)^\frac 12 d\theta$
$\int_0^{2\pi} (16 - 32 \cos\theta + 16 \cos^2\theta +16\sin^2\theta)^\frac 12 d\theta\\
\int_0^{2\pi} (32 - 32 \cos\theta)^\frac 12 d\theta\\
\int_0^{2\pi} 8(\frac{1 - \cos\theta}{2})^\frac 12 d\theta\\
\int_0^{2\pi} 8\sin\frac\theta2 d\theta\\
-16\cos\frac\theta2 |_0^{2\pi}\\
32$
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