Calculus of a Single Variable, Chapter 8, 8.3, Section 8.3, Problem 33

Separate the r and theta variables and integrate both sides:
int dr = int sin(pi*x)^4 dx
r + c_1=int sin(pi*x)^4 dx
Let pi*x=u, and pi*dx=du
r + c_1=int sin(u)^4 dx
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*x)
r+c_1=1/(8pi)[2pi*x-2sin(2pi*x)+pi*x+(1/4)sin(4pi*x)+c_2]
r+c_1=(2pi*x)/(8pi)-2sin(2pi*x)/(8pi)+(pi*x)/(8pi)+1/(32pi)sin(4pi*x)+c_2/(8pi)
Simplify terms and combine the constants into a new constant:
r=x/(4)-1/(4pi)sin(2pi*x)+x/8+1/(32pi)sin(4pi*x)+c
Your general solution is then:
r(x)=(3pi)/4-1/(4pi)sin(2pi*x)+1/(32pi)sin(4pi*x)+c