(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#
