Trig derivatives

  • #1

Homework Statement



[tex]\int[/tex]e[tex]^{2x}[/tex]sin(3x)dx

Homework Equations



integration by parts formula


The Attempt at a Solution



= 9/10e[tex]^{2x}[/tex](sin(3x)-1/3cos(3x))

I don't think I am taking the derivative of sin(3x) correctly

thanks for taking a look :)
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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[tex]\int e^{2x}sin(3x)dx[/tex]

u=e^2x so that du=2e^2x dx

dv=sin(3x)dx ; v=-1/3 cos(3x)


but either way you would have to do integration by parts twice so I am thinking that is not the answer(BUT I can be wrong as I did not really work it out)
 
  • #3
Hint :

What you do is you integrate it by parts two times.

A multiple of the original integral will appear in the answer.
 
  • #4
I tried it with u = e^2x ; dv = sin3xdx

I think i got the multiple and im stuck here:

= -1/3e[tex]^{2x}cos(3x) + 4/9[/tex][tex]\int[/tex]sin(3x)e[tex]^{2x}[/tex]
 
  • #5
rock.freak667
Homework Helper
6,223
31
I tried it with u = e^2x ; dv = sin3xdx

I think i got the multiple and im stuck here:

= -1/3e[tex]^{2x}cos(3x) + 4/9[/tex][tex]\int[/tex]sin(3x)e[tex]^{2x}[/tex]

you are supposed to get
[tex]\int e^{2x}sin(3x)dx=\frac{-e^{2x}}{3}cos(3x)+ \frac{2}{3}\int e^{2x}cos(3x)dx[/tex]

then do integration by parts again
 
  • #6
I got it, thanks guys

1/13 e^2x (2sin3x - 3cos3x) = integral
 

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