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

Separate the r and theta variables and integrate both sides:
int dr = int sin(pi*theta)^4 d(theta)
r + c_1=int sin(pi*theta)^4 d(theta)
Let pi*theta=u, and pi*d(theta)=du
r + c_1=int sin(u)^4 d(theta)
r + c_1=(1/pi)int sin(u)^4 du
r+c_1=(1/pi)int sin(u)^2*sin(u)^2 du
Use a trigonometric identity:
r+c_1=(1/pi) int (1/2)(1-cos(2u))(1/2)(1-cos(2u)) du
r+c_1=1/(4pi)int (1-2cos(2u)+cos(2u)^2) du
Let 2u=t, and 2du=dt
r+c_1=1/(4pi) int (1-2cos(t)+cos(t)^2)(dt/2)
r+c_1=1/(8pi) int (1-2cos(t)+(1/2)(1+cos(2t))) dt
r+c_1=1/(8pi)[int 1dt-2int cos(t) dt+(1/2)int dt+(1/2)int cos(2t) dt]
r+c_1=1/(8pi)[t-2sin(t)+1/2t+(1/2)(1/2)sin(2t)+c_2]
Substitute back in t=2u=2(pi*theta)
r+c_1=1/(8pi)[2pi*theta-2sin(2pi*theta)+pi*theta+(1/4)sin(4pi*theta)+c_2]
r+c_1=(2pi*theta)/(8pi)-2sin(2pi*theta)/(8pi)+(pi*theta)/(8pi)+1/(32pi)sin(4pi*theta)+c_2/(8pi)
Simplify terms and combine the constants c_1 and c_2 into a new constant c :
r=theta/(4)-1/(4pi)sin(2pi*theta)+theta/8+1/(32pi)sin(4pi*theta)+c
Your general solution is then:
r(theta)=(3pi)/4-1/(4pi)sin(2pi*theta)+1/(32pi)sin(4pi*theta)+c
Below is a plot from 0 to 2pi of a particular solution when c=0 .