Answer:
-30√(t/a) cos(√(at)) + 30/a sin(√(at)) + C
Explanation:
∫ 15 sin(√(at)) dt
Use substitution:
If x = √(at), then:
dx = ½ (at)^-½ (a dt)
dx = a / (2√(at)) dt
dx = a/(2x) dt
dt = (2/a) x dx
Plugging in:
∫ 15 sin x (2/a) x dx
30/a ∫ x sin x dx
Integrate by parts:
If u = x, then du = dx.
If dv = sin x dx, then v = -cos x.
∫ u dv = uv − ∫ v du
= 30/a (-x cos x − ∫ -cos x dx)
= 30/a (-x cos x + ∫ cos x dx)
= 30/a (-x cos x + sin x + C)
Substitute back:
30/a (-√(at) cos(√(at)) + sin(√(at)) + C)
-30√(t/a) cos(√(at)) + 30/a sin(√(at)) + C