We have to find the integral of sin (3x/2) * cos (x/2) dx
Int [ sin (3x/2) * cos (x/2) dx]
use sin 3x = 3sinx - 4(sinx)^3
=> Int [(3sin(x/2) - 4(sin(x/2))^3)*cos(x/2) dx]
let u = sin x/2
du/dx = (1/2)*cos (x/2)
=> 2*du = cos(x/2) dx
=> Int [(3u - 4u^3)*2 du]
=> 2* Int [(3u du] - 2* Int[ (4u^3) du]
=> 2* 3u^2 / 2 - 2* 4*u^4 /4 + C
=> 3u^2 - 2*u^4 + C
substitute u = sin x/2
=> 3*(sin x/2)^2 - 2*(sin x/2)^4 + C
=> (sin x/2)^2[ 3 - 2(sin x/2)^2] + C
=> (1/2)(1 - cos x)[ 2 + cos x] + C
=> (1/2)(2 - cos x - (cos x)^2) + C
=> (1/2)(2 - cos x - (1/2 + (1/2)(cos 2x)) + C
=> 1 - (1/2)cos x - 1/4 - (1/4)(cos 2x) + C
=> 3/4 - (1/4)(2cos x + (cos 2x)) + C
The 3/4 is added with the constant C
=> (-1/4)(cos 2x + 2*cos x) + C
The required integral is (-1/4)(cos 2x + 2*cos x) + C