Calculus of a Single Variable, Chapter 8, 8.3, Section 8.3, Problem 34

(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 .