anonymous
  • anonymous
How do you integrate e^2x sin(3x)dx?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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bahrom7893
  • bahrom7893
Use integration by parts. Btw when is this due?
anonymous
  • anonymous
its a revision, we cant figure it out
bahrom7893
  • 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
  • bahrom7893
ill just do it now
anonymous
  • anonymous
we would just like to know what to do with the sin(3x)
anonymous
  • anonymous
thanks!
bahrom7893
  • bahrom7893
ima write this out on paper and email it to u. what's ur email?
anonymous
  • anonymous
moca341@hotmail.com .. thank you :)
bahrom7893
  • bahrom7893
np
bahrom7893
  • bahrom7893
almost done...
anonymous
  • anonymous
great!
bahrom7893
  • bahrom7893
i will take separate pix and email them one by one, the whole solution won't fit lol
anonymous
  • anonymous
thats fine! haha
bahrom7893
  • bahrom7893
okay emailed it
bahrom7893
  • bahrom7893
ask me if u have any questions, either here or by email and Fan if I helped!
bahrom7893
  • bahrom7893
did u get it?
anonymous
  • anonymous
It helped a lot, but i dont really get it!, ill try to figure it out! thanks a lot!
bahrom7893
  • bahrom7893
no, ask me which part don't you understand?
bahrom7893
  • bahrom7893
I think it might the place where I let that one integral be capital i
anonymous
  • anonymous
i dont even understand the first line, dont worry about it!
bahrom7893
  • 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
  • anonymous
i dont see where the 1/3 cos (3x) comes from
bahrom7893
  • bahrom7893
Oh okay that's just the integral of Sin(3x)
bahrom7893
  • 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
  • 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
  • anonymous
ok, i get it, thanks a lot!
bahrom7893
  • bahrom7893
\[v = \int\limits_{}^{}Sin(3x)*(3/3)*dx = (1/3) *\int\limits_{}^{}Sin(3x)*3*dx\]
bahrom7893
  • bahrom7893
Now you have both a and da; your integral simplifies to: \[(1/3)\int\limits_{}^{}Sin(a)*da\]
bahrom7893
  • bahrom7893
that's it then integrate as u would a regular sin and in the end replace a by 3x

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