Answer
$$r =\frac{1}{4}\left(\frac{3\theta}{2}-\frac{1}{\pi}\sin 2\pi \theta+\frac{1}{8\pi}\sin 4\pi \theta\right) +c$$
Work Step by Step
$$
\frac{d r}{d \theta}=\sin ^{4} \pi \theta
$$
Since $$\sin^2x=\frac{1}{2}(1-\cos 2x) ,\ \ \ \cos^2x=\frac{1}{2}(1+\cos 2x)$$
Then
\begin{align*}
dr&= \sin ^{4} \pi \theta d\theta \\
r&=\int \sin ^{4} \pi \theta d\theta\\
&=\frac{1}{4}\int(1-\cos 2\pi \theta) ^2d\theta\\
&=\frac{1}{4}\int(1-2\cos 2\pi \theta+\cos^2 2\pi \theta) d\theta\\
&=\frac{1}{4}\int\left(1-2\cos 2\pi \theta+\frac{1}{2}+\frac{1}{2}\cos 4\pi \theta\right) d\theta\\
&=\frac{1}{4}\int\left(\frac{3}{2}-2\cos 2\pi \theta+\frac{1}{2}\cos 4\pi \theta\right)d\theta\\
&=\frac{1}{4}\left(\frac{3\theta}{2}-\frac{1}{\pi}\sin 2\pi \theta+\frac{1}{8\pi}\sin 4\pi \theta\right) +c\\
\end{align*}
