Integrate ∫ sin3x sin4x dx

Note that the "sin3x" and "sin4x" are not powers.
Please show and explain any steps that you take.

The best first step is to use the following neat trig identity:

sinAsinB = (1/2)(cos(A - B) - cos(A + B))

from that, you have

(1/2)∫[cos(x) - cos(7x)]dx

from there, you can separate into a simple integral and a u-substitution with u = 7x.

End result:

(1/2)sin(x) - (1/14)sin(7 x) + C

There are 2 ways we can solve this: integration by parts, or we could use a trig identity

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

sin(3x) * sin(4x) * dx =>
(1/2) * (cos(3x - 4x) - cos(3x + 4x)) * dx =>
(1/2) * (cos(-x) - cos(7x)) * dx

Integrate:

(1/2) * (-sin(-x) - (1/7) * sin(7x)) + C =>
(1/2) * (sin(x) - (1/7) * sin(7x)) + C =>
(1/14) * (7sin(x) - sin(7x)) + C


Integrate by parts:

u = sin(3x)
du = 3 * cos(3x) * dx
dv = sin(4x) * dx
v = (-1/4) * cos(4x)

int(u * dv) =>
uv - int(v * du) =>
(-1/4) * sin(3x) * cos(4x) + (3/4) * int(cos(3x) * cos(4x) * dx)

u = cos(3x)
du = -3 * sin(3x) * dx
dv = cos(4x) * dx
v = (1/4) * sin(4x)

(-1/4) * sin(3x) * cos(4x) + (3/4) * ((1/4) * sin(4x) * cos(3x) + (3/4) * int(sin(4x) * sin(3x) * dx)) =>
(-1/4) * sin(3x) * cos(4x) + (3/16) * sin(4x) * cos(3x) + (9/16) * int(sin(4x) * sin(3x) * dx)

Now we have:

int(sin(4x) * sin(3x) * dx) = (-1/4) * sin(3x) * cos(4x) + (3/16) * sin(4x) * cos(3x) + (9/16) * int(sin(4x) * sin(3x) * dx)
int(sin(4x) * sin(3x) * dx) - (9/16) * int(sin(4x) * sin(3x) * dx) = (3/16) * sin(4x) * cos(3x) - (1/4) * sin(3x) * cos(4x)
int(sin(4x) * sin(3x) * dx) * (1 - (9/16)) = (3/16) * sin(4x) * cos(3x) - (1/4) * sin(3x) * cos(4x)
int(sin(4x) * sin(3x) * dx) * (7/16) = (3/16) * sin(4x) * cos(3x) - (1/4) * sin(3x) * cos(4x)
int(sin(4x) * sin(3x) * dx) = (16/7) * (1/16) * (3 * sin(4x) * cos(3x) - 4 * sin(3x) * cos(4x))
int(sin(4x) * sin(3x) * dx) = (1/7) * (3 * sin(4x) * cos(3x) - 4 * sin(3x) * cos(4x))

Add the constant of integration

int(sin(4x) * sin(3x) * dx) = (1/7) * (3 * sin(4x) * cos(3x) - 4 * sin(3x) * cos(4x)) + C