How do you integrate #int e^(3x)cos(2x)# by integration by parts method?

1 Answer
Gió · Steve M
Feb 2, 2017

I got: #9/13[e^(3x)/3cos(2x)+2/9e^(3x)sin(2x)]+c#
BUT check my maths.

Explanation:

Ok let us try:
#inte^(3x)cos(2x)dx=e^(3x)/3cos(2x)-int(e^(3x)/3)(-2sin(2x))dx=e^(3x)/3cos(2x)+2/3int(e^(3x))sin(2x)dx=#
again:
#=e^(3x)/3cos(2x)+2/3[e^(3x)/3sin(2x)-int(e^(3x)/3)(2cos(2x))dx]#

So basically we get:

#inte^(3x)cos(2x)dx=e^(3x)/3cos(2x)+2/9e^(3x)sin(2x)-4/9inte^(3x)(2cos(2x))dx#

Now a trick....
Let us take #-4/9inte^(3x)(2cos(2x))dx# to the left of the #=# sign:
#inte^(3x)cos(2x)dx+4/9inte^(3x)(2cos(2x))dx=e^(3x)/3cos(2x)+2/9e^(3x)sin(2x)#
add the two integral on the left:
#13/9inte^(3x)(2cos(2x))dx=e^(3x)/3cos(2x)+2/9e^(3x)sin(2x)#
and:
#inte^(3x)(2cos(2x))dx=9/13[e^(3x)/3cos(2x)+2/9e^(3x)sin(2x)]+c#

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