# `(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  ...

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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` .

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