Hi there!

To find the indefinite integral, we must integrate by parts.
Let "u" be the expression most easily differentiated, and "dv" the remaining expression. Take the derivative of "u" and the integral of "dv":
u = 4x
du = 4
dv = cos(2 - 3x)
v = 1/3sin(2 - 3x)
Write into the format:
∫udv = uv - ∫vdu
Thus, utilize the solved for expressions above:
4x · (-1/3sin(2 - 3x)) -∫ 4(1/3sin(2 - 3x))dx
Simplify:
-4x/3 sin(2 - 3x) - ∫ 4/3sin(2 - 3x)dx
Integrate the integral:
∫4/3(sin(2 - 3x)dx
u = 2 - 3x
du = -3dx ⇒ -1/3du = dx
-1/3∫ 4/3(sin(2 - 3x)dx ⇒ -4/9cos(2 - 3x) + C
Combine:

F(-4) = 6, f(0) = -6, f(1) = -4
SohcahtoaMy algebra teacher got us all to remember that word because...
Soh -
Sin =
Opposite /
Hypotenus
Cah -
Cos =
Adjacent /
Hypotenus
Toa -
Tan =
Opposite /
Adjacent
Sin, Cos, and Tan are what you can use to find missing triangle sides if you know at least one other angle besides the 90 degree angle.