(1) \inte^(8x)sin(9x) dx
We need to use integration by parts.
\int udv = uv - \intvdu
It's useful to remember the acronym LIATE: Log, Inverse Trig, Algebraic, Trig, Exponential for knowing the prioritization of choosing your u value.
Since trigonometric functions take precedence over exponentials, we will use u = sin(9x).
Substitution:
u= sin(9x)
du = 9cos(9x)dx
dv = e^(8x)dx
v = 1/8e^(8x)
(2) \inte^(8x)sin(9x) dx = 1/8e^(8x)sin(9x)-9/8\inte^(8x)cos(9x)dx
Since we haven't improved our situation, it looks like another round of integration by parts. Let's use the letters y and z, and let y = cos(9x) and integrate the RHS integral by parts.
\int ydz = yz - \intzdy
Substitution:
y = cos(9x)
dy = -9sin(9x)dx
dz = e^(8x)dx
z = 1/8e^(8x)
(3) \inte^(8x)cos(9x)dx = 1/8e^(8x)cos(9x)+9/8\inte^(8x)sin(9x)dx
Substituting (3) -> (2):
(4) \inte^(8x)sin(9x) dx = 1/8e^(8x)sin(9x) - 9/8 (1/8e^(8x)cos(9x)+9/8\inte^(8x)sin(9x)dx)
\inte^(8x)sin(9x) dx = 1/8e^(8x)sin(9x) - 9/64 e^(8x)cos(9x)-81/64\inte^(8x)sin(9x)dx
145/64\inte^(8x)sin(9x) dx = 1/8e^(8x)sin(9x) - 9/64 e^(8x)cos(9x)
= 64/145(1/8e^(8x)sin(9x) - 9/64 e^(8x)cos(9x)) +C
= 8/145e^(8x)sin(9x) - 9/145 e^(8x)cos(9x) +C
=1/145e^(8x)(8sin(9x)-9cos(9x)) +C
