Evaluate: ∫sin(x−a)sin(x+a)dx

Let I = ∫sin(x−a)sin(x+a)dx
Put x+a = t, we have dx = dt
I = ∫sin(t−2a)sint dt
I = ∫(sintcos2a − cost sin2a)sint dt
I = cos2a∫sint dt − sin2a∫sint cost dt
I = cos2a∫sint dt − sin2a∫(1/2)sin2t dt
I = cos2a(-cost) - (sin2a/2)(-cos2t)/2 + C
I = -cost cos2a + (sin2a/4)cos2t + C
I = -cos(x+a)cos2a + (sin2a/4)cos(2(x+a)) + C
Alternatively:
I = cos2a∫sin²t dt - sin2a∫sin t cos t dt
Using sin²t = (1-cos2t)/2 and sin t cos t = (1/2)sin2t,
I = cos2a∫(1-cos2t)/2 dt - (sin2a/2)∫sin2t dt
I = (cos2a/2)∫(1-cos2t)dt - (sin2a/2)∫sin2t dt
I = (cos2a/2)(t - (sin2t)/2) - (sin2a/2)(-cos2t)/2 + C
I = (cos2a/2)(t - (sin2t)/2) + (sin2a/4)cos2t + C
I = (cos2a/2)(x+a - (sin(2(x+a)))/2) + (sin2a/4)cos(2(x+a)) + C
I = (x+a)cos2a/2 - (cos2a/4)sin(2(x+a)) + (sin2a/4)cos(2(x+a)) + C
Another approach:
Using the product-to-sum formula:
2sin(x-a)sin(x+a) = cos(2a) - cos(2x)
∫sin(x-a)sin(x+a)dx = (1/2)∫[cos(2a) - cos(2x)]dx
= (1/2)[xcos(2a) - (sin(2x))/2] + C
= xcos(2a)/2 - sin(2x)/4 + C