Ayay stuck on integral by parts problem.

[Alpha6]

Please help.


The term I have to integrate is: e^3xcos(6x)

∫e^3xcos(6x)dx uv - ∫ vdu

u = e^3x du = 3e^3x

dv = cos(6x) v = sin6x/6


Therefore:

e3x (sin6x/6) - ∫ (sin6x/6)3e^3x dx

^{↓ Simplify}

e3x sin6x/6 - ∫ (sin6xe^3x/2 dx


↓ pull out the 1/2



e^3x sin6x/6 - 1/2∫ sin6xe^3xdx


↓ integrate again. u = e^3x du = 3e^3x

dv = sin6x v = - cos6x/6

-1/2 (e^3x - cos6x/6 - ∫ - cos6x/6 (3e^3x)

↓ double negatives

-1/2 (e^3x - cos6x/6 + ∫ cos6x/6 (3e^3x)

↓ Simplify


-1/2 (e^3x - cos6x/6 + ∫ cos6x (e^3x)/2


Okay. I get stuck here because it seems like I'm just doing an endless cycle of u v - ∫ v u'

So I peeked at some online solutions and it ended up being : e^3x(2sin(6x) + cos (6x)) / 15

I can't for the life of me figure out how they arrived at that answer. I am pretty sure I am doing this right any help??

 

[Deleted member 4993]

Please help.


The term I have to integrate is: e^3xcos(6x)

∫e^3xcos(6x)dx uv - ∫ vdu

u = e^3x du = 3e^3x

dv = cos(6x) v = sin6x/6


Therefore:

e3x (sin6x/6) - ∫ (sin6x/6)3e^3x dx

^{↓ Simplify}

e3x sin6x/6 - ∫ (sin6xe^3x/2 dx


↓ pull out the 1/2



e^3x sin6x/6 - 1/2∫ sin6xe^3xdx


↓ integrate again. u = e^3x du = 3e^3x

dv = sin6x v = - cos6x/6

-1/2 (e^3x - cos6x/6 - ∫ - cos6x/6 (3e^3x)

↓ double negatives

-1/2 (e^3x - cos6x/6 + ∫ cos6x/6 (3e^3x)

↓ Simplify


-1/2 (e^3x - cos6x/6 + ∫ cos6x (e^3x)/2


Okay. I get stuck here because it seems like I'm just doing an endless cycle of u v - ∫ v u'

So I peeked at some online solutions and it ended up being : e^3x(2sin(6x) + cos (6x)) / 15

I can't for the life of me figure out how they arrived at that answer. I am pretty sure I am doing this right any help??

What you got is

I = A*I + B

then

I = B/(1-A)

 

[Ishuda]

OK - the way I would approach this is to note that
cos(x) = \(\displaystyle \frac{e^{i x} +\space e^{-i x}}{2}\)
sin(x) = \(\displaystyle \frac{e^{i x} -\space e^{-i x}}{2i}\)

So, first, split the cosine to do the integration and then put the cosine and sine back together. Oh, note that -i = 1/i.