Trig derivatives

1. Jan 15, 2008

Wonderballs

1. The problem statement, all variables and given/known data

$$\int$$e$$^{2x}$$sin(3x)dx

2. Relevant equations

integration by parts formula

3. The attempt at a solution

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

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

thanks for taking a look :)

2. Jan 15, 2008

rock.freak667

$$\int e^{2x}sin(3x)dx$$

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. Jan 15, 2008

silver-rose

Hint :

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

A multiple of the original integral will appear in the answer.

4. Jan 15, 2008

Wonderballs

I tried it with u = e^2x ; dv = sin3xdx

I think i got the multiple and im stuck here:

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

5. Jan 15, 2008

rock.freak667

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

then do integration by parts again

6. Jan 15, 2008

Wonderballs

I got it, thanks guys

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