Notes: `(dr)/(d theta) = sin^4(pitheta)` Solve the differential equation.

Separate the r and $\displaystyle\theta$ variables and integrate both sides:

$\displaystyle\int{d}{r}=\int{{\sin{{\left(\pi\cdot\theta\right)}}}}^{{4}}{d}{\left(\theta\right)}$

$\displaystyle{r}+{c}_{{1}}=\int{{\sin{{\left(\pi\cdot\theta\right)}}}}^{{4}}{d}{\left(\theta\right)}$

Let $\displaystyle\pi\cdot\theta={u},{\quad\text{and}\quad}\pi\cdot{d}{\left(\theta\right)}={d}{u}$

$\displaystyle{r}+{c}_{{1}}=\int{{\sin{{\left({u}\right)}}}}^{{4}}{d}{\left(\theta\right)}$

$\displaystyle{r}+{c}_{{1}}={\left(\frac{{1}}{\pi}\right)}\int{{\sin{{\left({u}\right)}}}}^{{4}}{d}{u}$

$\displaystyle{r}+{c}_{{1}}={\left(\frac{{1}}{\pi}\right)}\int{{\sin{{\left({u}\right)}}}}^{{2}}\cdot{{\sin{{\left({u}\right)}}}}^{{2}}{d}{u}$

Use a trigonometric identity:

$\displaystyle{r}+{c}_{{1}}={\left(\frac{{1}}{\pi}\right)}\int{\left(\frac{{1}}{{2}}\right)}{\left({1}-{\cos{{\left({2}{u}\right)}}}\right)}{\left(\frac{{1}}{{2}}\right)}{\left({1}-{\cos{{\left({2}{u}\right)}}}\right)}{d}{u}$

$\displaystyle{r}+{c}_{{1}}=\frac{{1}}{{{4}\pi}}\int{\left({1}-{2}{\cos{{\left({2}{u}\right)}}}+{{\cos{{\left({2}{u}\right)}}}}^{{2}}\right)}{d}{u}$

Let $\displaystyle{2}{u}={t},{\quad\text{and}\quad}{2}{d}{u}={\left.{d}{t}\right.}$

$\displaystyle{r}+{c}_{{1}}=\frac{{1}}{{{4}\pi}}\int{\left({1}-{2}{\cos{{\left({t}\right)}}}+{{\cos{{\left({t}\right)}}}}^{{2}}\right)}{\left(\frac{{\left.{d}{t}}}{{2}}\right)}$

$\displaystyle{r}+{c}_{{1}}=\frac{{1}}{{{8}\pi}}\int{\left({1}-{2}{\cos{{\left({t}\right)}}}+{\left(\frac{{1}}{{2}}\right)}{\left({1}+{\cos{{\left({2}{t}\right)}}}\right)}\right)}{\left.{d}{t}\right.}$

$\displaystyle{r}+{c}_{{1}}=\frac{{1}}{{{8}\pi}}{\left[\int{1}{\left.{d}{t}\right.}-{2}\int{\cos{{\left({t}\right)}}}{\left.{d}{t}\right.}+{\left(\frac{{1}}{{2}}\right)}\int{\left.{d}{t}\right.}+{\left(\frac{{1}}{{2}}\right)}\int{\cos{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}\right]}$

$\displaystyle{r}+{c}_{{1}}=\frac{{1}}{{{8}\pi}}{\left[{t}-{2}{\sin{{\left({t}\right)}}}+\frac{{1}}{{2}}{t}+{\left(\frac{{1}}{{2}}\right)}{\left(\frac{{1}}{{2}}\right)}{\sin{{\left({2}{t}\right)}}}+{c}_{{2}}\right]}$

Substitute back in $\displaystyle{t}={2}{u}={2}{\left(\pi\cdot\theta\right)}$

$\displaystyle{r}+{c}_{{1}}=\frac{{1}}{{{8}\pi}}{\left[{2}\pi\cdot\theta-{2}{\sin{{\left({2}\pi\cdot\theta\right)}}}+\pi\cdot\theta+{\left(\frac{{1}}{{4}}\right)}{\sin{{\left({4}\pi\cdot\theta\right)}}}+{c}_{{2}}\right]}$

$\displaystyle{r}+{c}_{{1}}=\frac{{{2}\pi\cdot\theta}}{{{8}\pi}}-{2}\frac{{\sin{{\left({2}\pi\cdot\theta\right)}}}}{{{8}\pi}}+\frac{{\pi\cdot\theta}}{{{8}\pi}}+\frac{{1}}{{{32}\pi}}{\sin{{\left({4}\pi\cdot\theta\right)}}}+\frac{{c}_{{2}}}{{{8}\pi}}$

Simplify...

Separate the r and $\displaystyle\theta$ variables and integrate both sides:

$\displaystyle\int{d}{r}=\int{{\sin{{\left(\pi\cdot\theta\right)}}}}^{{4}}{d}{\left(\theta\right)}$

$\displaystyle{r}+{c}_{{1}}=\int{{\sin{{\left(\pi\cdot\theta\right)}}}}^{{4}}{d}{\left(\theta\right)}$

Let $\displaystyle\pi\cdot\theta={u},{\quad\text{and}\quad}\pi\cdot{d}{\left(\theta\right)}={d}{u}$

$\displaystyle{r}+{c}_{{1}}=\int{{\sin{{\left({u}\right)}}}}^{{4}}{d}{\left(\theta\right)}$

$\displaystyle{r}+{c}_{{1}}={\left(\frac{{1}}{\pi}\right)}\int{{\sin{{\left({u}\right)}}}}^{{4}}{d}{u}$

$\displaystyle{r}+{c}_{{1}}={\left(\frac{{1}}{\pi}\right)}\int{{\sin{{\left({u}\right)}}}}^{{2}}\cdot{{\sin{{\left({u}\right)}}}}^{{2}}{d}{u}$

Use a trigonometric identity:

$\displaystyle{r}+{c}_{{1}}=\frac{{1}}{{{4}\pi}}\int{\left({1}-{2}{\cos{{\left({2}{u}\right)}}}+{{\cos{{\left({2}{u}\right)}}}}^{{2}}\right)}{d}{u}$

Let $\displaystyle{2}{u}={t},{\quad\text{and}\quad}{2}{d}{u}={\left.{d}{t}\right.}$

$\displaystyle{r}+{c}_{{1}}=\frac{{1}}{{{4}\pi}}\int{\left({1}-{2}{\cos{{\left({t}\right)}}}+{{\cos{{\left({t}\right)}}}}^{{2}}\right)}{\left(\frac{{\left.{d}{t}}}{{2}}\right)}$

Substitute back in $\displaystyle{t}={2}{u}={2}{\left(\pi\cdot\theta\right)}$

Simplify terms and combine the constants $\displaystyle{c}_{{1}}$ and $\displaystyle{c}_{{2}}$ into a new constant $\displaystyle{c}$ :

$\displaystyle{r}=\frac{\theta}{{{4}}}-\frac{{1}}{{{4}\pi}}{\sin{{\left({2}\pi\cdot\theta\right)}}}+\frac{\theta}{{8}}+\frac{{1}}{{{32}\pi}}{\sin{{\left({4}\pi\cdot\theta\right)}}}+{c}$

Your general solution is then:

$\displaystyle{r}{\left(\theta\right)}=\frac{{{3}\pi}}{{4}}-\frac{{1}}{{{4}\pi}}{\sin{{\left({2}\pi\cdot\theta\right)}}}+\frac{{1}}{{{32}\pi}}{\sin{{\left({4}\pi\cdot\theta\right)}}}+{c}$