Solving Trig Derivatives with Integration by Parts

[Wonderballs]

Homework Statement

\inte^{2x}sin(3x)dx

Homework Equations

integration by parts formula

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 :)

 

[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)

 

[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.

 

[Wonderballs]

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

I think i got the multiple and I am stuck here:

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

 

[rock.freak667]

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

I think i got the multiple and I am stuck here:

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

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

 

[Wonderballs]

I got it, thanks guys

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