`(ds)/(d alpha) = sin^2(alpha/2)cos^2(alpha/2)`

To solve, express the differential equation in the form N(y)dy = M(x)dx .

So bringing together same variables on one side, the equation becomes

`ds =sin^2(alpha/2)cos^2(alpha/2) d alpha`

To simplify the right side, apply the exponent rule `(ab)^m=a^mb^m` .

`ds =(sin(alpha/2)cos(alpha/2))^2 d alpha`

Then, apply the sine double angle identity `sin(2 theta)=2sin(theta)cos(theta)` .

`sin (2*alpha/2)=2sin(alpha/2)cos(alpha/2)`

`sin(alpha)=2sin(alpha/2)cos(alpha/2)`

`sin(alpha)=2sin(alpha/2)cos(alpha/2)`

`(sin(alpha))/2=sin(alpha/2)cos(alpha/2)`

Substituting this to the right side, the differential equation becomes

`ds = ((sin (alpha))/2)^2 d alpha`

`ds = (sin^2 (alpha))/4 d alpha`

Then, apply the cosine double angle identity  `cos(2 theta)=1-2sin^2(theta)` .

`cos (2alpha) = 1 - 2sin^2(alpha)`

`2sin^2(alpha) = 1-cos(2 alpha)`

`sin^2(alpha) = (1-cos(2 alpha))/2`

Plugging this to the right side, the differential equation becomes

`ds = ((1-cos(2 alpha))/2)/4 d alpha`

`ds = (1-cos(2alpha))/8 d alpha`

`ds = (1/8 - cos(2alpha)/8) d alpha`

Then, take the integral of both sides.

`int ds = int (1/8 - cos(2alpha)/8) d alpha`

`int ds = int 1/8 d alpha - int cos(2 alpha)/8 d alpha`

Apply the integral formulas `int adx = ax + C` and `int cos(x) dx = sin(x) + C` .

`s+C_1 = 1/8alpha - (sin(2alpha))/16 + C_2`

Then, isolate the s.

`s = 1/8alpha - (sin(2alpha))/16 + C_2-C_1`

Since C1 and C2 represents any number, it can be expressed as a single constant C.

`s = 1/8alpha - (sin(2alpha))/16 +C`

Therefore, the general solution of the differential equation is `s = 1/8alpha - (sin(2alpha))/16 +C` .