integral of sin(3x)cos(2x)dx?

Trigonometric identities used :
sin(3x) = -4 sin³(x) + 3 sin(x)
cos(2x) = 2 cos²(x) - 1
sin²(x) + cos²(x) = 1, and thus sin²(x) = 1 - cos²(x)

∫ sin(3x) cos(2x) dx
= ∫ [-4 sin³(x) + 3 sin(x)] [2 cos²(x) - 1] dx
= ∫ [-4 sin²(x) sin(x) + 3 sin(x)] [2 cos²(x) - 1] dx
= ∫ {-4 [1 - cos²(x)] sin(x) + 3 sin(x)} [2 cos²(x) - 1] dx
= ∫ [-4 sin(x) + 4 cos²(x) sin(x) + 3 sin(x)] [2 cos²(x) - 1] dx
= ∫ [4 cos²(x) sin(x) - sin(x)] [2 cos²(x) - 1] dx
= ∫ sin(x) [4 cos²(x) - 1] [2 cos²(x) - 1] dx
= ∫ [4 cos²(x) - 1] [2 cos²(x) - 1] [sin(x) dx]
= ∫ [8 cos⁴(x) - 6 cos²(x) + 1] [d cos(x)]
= ∫8 cos⁴(x) [d cos(x)] - ∫ 6 cos²(x) [d cos(x)] + ∫ d cos(x)
= 8 * (1/5) * cos⁵(x) - 6 * (1/3) * cos³(x) + sin(x) + C
= (8/5) cos⁵(x) - 2 cos³(x) + sin(x) + C

sin(3x)cos(2x)=
(1/2)[2sin(3x)cos(2x)]=
(1/2)[sin(5x)+sin(x)]
(product to sum formula)
=>
Ssin(3x)cos(2x)dx
=
(1/2)[Ssin(5x)dx+Ssin(x)dx]+C
=
(1/2)[-cos(5x)/5-cos(x)]+C
=
C-cos(5x)/10-cos(x)/2

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There are a few ways you can do this. A straightforward way involves integration by parts, twice

u = sin(3x)
du = 3 * cos(3x) * dx
dv = cos(2x) * dx
v = (1/2) * sin(2x)

int(u * dv) = u * v - int(v * du)

int(sin(3x) * cos(2x) * dx) = (1/2) * sin(2x) * sin(3x) - (3/2) * int(cos(3x) * sin(2x) * dx)

u = cos(3x)
du = -3 * sin(3x) * dx
dv = sin(2x) * dx
v = (-1/2) * cos(2x)

int(sin(3x) * cos(2x) * dx) = (1/2) * sin(2x) * sin(3x) - (3/2) * ((-1/2) * cos(2x) * cos(3x) - (-3/-2) * int(sin(3x) * cos(2x) * dx))

int(sin(3x) * cos(2x) * dx) = t

t = (1/2) * sin(2x) * sin(3x) - (3/2) * ((-1/2) * cos(2x) * cos(3x) - (3/2) * t)
t = (1/2) * sin(2x) * sin(3x) + (3/4) * cos(2x) * cos(3x) + (9/4) * t
t - (9/4) * t = (1/4) * (2 * sin(2x) * sin(3x) + 3 * cos(2x) * cos(3x))
(1 - 9/4) * t = (1/4) * (2 * sin(2x) * sin(3x) + 3 * cos(2x) * cos(3x))
(-5/4) * t = (1/4) * (2 * sin(2x) * sin(3x) + 3 * cos(2x) * cos(3x))
t = (-1/5) * (2 * sin(2x) * sin(3x) + 3 * cos(2x)* cos(3x))


(-1/5) * (2 * sin(2x) * sin(3x) + 3 * cos(2x) * cos(3x))

Add the constant of integration

(-1/5) * (2 * sin(2x) * sin(3x) + 3 * cos(2x) * cos(3x)) + C


the other method involves a product-to-sum identity.

sin(a) * cos(b) = (1/2) * (sin(a + b) + sin(a - b))

sin(3x) * cos(2x) * dx =>
(1/2) * (sin(3x + 2x) + sin(3x - 2x)) * dx =>
(1/2) * (sin(5x) + sin(x)) * dx

Integrate

(1/2) * ((-1/5) * cos(5x) - cos(x)) + C =>
(-1/2) * (1/5) * (cos(5x) + 5 * cos(x)) + C =>
(-1/10) * (cos(5x) + 5 * cos(x)) + C

We can compare it to our previous answer

(-1/5) * (2 * sin(2x) * sin(3x) + 3 * cos(2x) * cos(3x)) + C
(-1/5) * (2 * (1/2) * (cos(2x - 3x) - cos(2x + 3x)) + 3 * (1/2) * (cos(2x + 3x) + cos(2x - 3x))) + C
(-1/5) * (cos(-x) - cos(5x) + (3/2) * cos(5x) + (3/2) * cos(-x)) + C =>
(-1/5) * (1/2) * (2 * cos(-x) - 2 * cos(5x) + 3 * cos(5x) + 3 * cos(-x)) + C =>
(-1/10) * (2cos(x) + 3cos(x) + cos(5x)) + C =>
(-1/10) * (5 * cos(x) + cos(5x)) + C

Same answer.

sin(3x + 2x) = sin(3x) cos(2x) + cos(3x)sin(2x)
sin(3x - 2x) = sin(3x) cos(2x) - cos(3x)sin(2x)
sin(3x)cos(2x) = (1/2)[sin(5x) + sin(x)]

Let I = ∫[sin(3x)cos(2x)] dx
I = -[(1/10)cos(5x) + (1/2)cos(x)] + C