How do you find the integral of #f(x)=e^(2x)sin3x# using integration by parts?

1 Answer
Nov 5, 2015

#1/13e^(2x)(2sin(3x) - 3cos(3x)) + c#

Explanation:

So, with integration by parts we state that:

#intudv = uv - intvdu#

Let's use #e^(2x)# as our u function and #sin3x# as our v' function, leaving us with:

#u = e^(2x)#
#u' = 2e^(2x)#

#v = -1/3cos3x#
#v' = sin3x#

Now let's plug this back into our original formula:

#e^(2x)(-1/3cos3x) - int2e^(2x)(-1/3cos3x)dx#

Let's simplify it a bit more:

#-1/3e^(2x)(cos3x) - (-2/3)inte^(2x)(cos3x)dx#

We have to apply integration by parts yet again, using:

#u = e^(2x)#
#u' = 2e^(2x)#

#v'=cos(3x)#
#v=1/3sin(3x)#

Ultimately giving us:

#-1/3e^(2x)cos(3x)-(-2/3(e^(2x)1/3sin(3x)-int2e^(2x)1/3sin(3x)dx))#

Now, if we isolate #intsin(3x)e^(2x)dx# from this equation we'll end up with:

#intsin(3x)e^(2x)dx = 1/13e^(2x)(2sin(3x)-3cos(3x)) + C#

I'm taking Calc 2 as well, so if anyone has a correction please don't hesitate! Also I'm running late for class so I kinda glossed over the last step, I can explain further if needed!