How do you find the integral of #(e^(8x))sin(9x)dx#?

Question

Luke Alvarez · Accepted Answer

#\inte^(8x)sin(9x) dx=1/145e^(8x)(8sin(9x)-9cos(9x)) +C#

(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#

Aurora Alvarado · Answer

undefined To find the integral of $ e^{8x} \sin(9x) $ with respect to $ x $, you can use integration by parts. Let $ u = e^{8x} $ and $ dv = \sin(9x)dx $. Then, differentiate $ u $ to find $ du $, and integrate $ dv $ to find $ v $. Afterward, apply the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

By substituting the expressions for $ u $, $ dv $, $ du $, and $ v $, you can solve for the integral.