How do you integrate e^2x sin(3x)dx?

- anonymous

How do you integrate e^2x sin(3x)dx?

- katieb

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- bahrom7893

Use integration by parts. Btw when is this due?

- anonymous

its a revision, we cant figure it out

- bahrom7893

No i just meant i have to do my own hw too, so do u need this like right away or by tonite or by 2morro mornin?

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- bahrom7893

ill just do it now

- anonymous

we would just like to know what to do with the sin(3x)

- anonymous

thanks!

- bahrom7893

ima write this out on paper and email it to u. what's ur email?

- anonymous

moca341@hotmail.com .. thank you :)

- bahrom7893

np

- bahrom7893

almost done...

- anonymous

great!

- bahrom7893

i will take separate pix and email them one by one, the whole solution won't fit lol

- anonymous

thats fine! haha

- bahrom7893

okay emailed it

- bahrom7893

ask me if u have any questions, either here or by email and Fan if I helped!

- bahrom7893

did u get it?

- anonymous

It helped a lot, but i dont really get it!, ill try to figure it out!
thanks a lot!

- bahrom7893

no, ask me which part don't you understand?

- bahrom7893

I think it might the place where I let that one integral be capital i

- anonymous

i dont even understand the first line, dont worry about it!

- bahrom7893

Oh okay so first I used integration by parts. See the stuff in the circle:
I said Let u = e^(2x), then what is du/dx?
du/dx = derivative of u = 2e^(2x), multiply both sides by dx to get:
du = 2e^(2x)dx

- anonymous

i dont see where the 1/3 cos (3x) comes from

- bahrom7893

Oh okay that's just the integral of Sin(3x)

- bahrom7893

I will work it out here:
After that I said let dv/dx = Sin(3x), well then what's V? v is the integral of dv/dx with respect to x.
V = Integral(dv/dx, dx)
\[v = \int\limits_{}^{}Sin(3x)dx\]

- bahrom7893

Here you have to use a simple substitution:
Let a = 3x; then da = 3 dx. Now you have 3x in your integral, but you still need a 3 in front of dx. To do so, multiply and divide by 3, same as multiplying by 1

- anonymous

ok, i get it, thanks a lot!

- bahrom7893

\[v = \int\limits_{}^{}Sin(3x)*(3/3)*dx = (1/3) *\int\limits_{}^{}Sin(3x)*3*dx\]

- bahrom7893

Now you have both a and da; your integral simplifies to:
\[(1/3)\int\limits_{}^{}Sin(a)*da\]

- bahrom7893

that's it then integrate as u would a regular sin and in the end replace a by 3x

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