Calculus integration problem

[Joker2k999]

Good day guys,
Having a hard time trying to integrate this prob.

∫e^-3t sin wt dt

i believe its integration by parts but not sure.

thx in advance.
JkR.

 

[Joker2k999]

i have my u= sin (wt) v = -1/3e^-3t
du= cos wt/w dv =e^-3t

uv-∫ vdu

-1/3∫e^-3t * cos wt/w

 

[renegade05]

You're going to have to use integration by parts twice. I also like to set my integral to a variable, say a = ∫e^-3t sin wt dt

let
\(\displaystyle u=e^{-3t}\)
\(\displaystyle du=-3e^{-3t}\)
\(\displaystyle dv=\sin{t}\)
\(\displaystyle v=-\cos{t}\)

I am guessing the "w" in your sin wt is a typo? If not, easy enough, just treat as a constant -
\(\displaystyle dv=\sin{wt}\)

\(\displaystyle v=-\frac{\cos{wt}}{w}\)

So now use integration by parts (a = uv - ∫vdu) , and then use it again on the integral you get from your ∫vdu, and you should get your original integral back (∫e^-3t sin wt dt) which of course is your variable "a". So now solve for a and that's what the integral equals. Hope this helps. Im sure one of the hot shots on the forum will correct any errors I made, but this should get you started.

 

[galactus]

Let's use parts to derive the general form \(\displaystyle \int e^{at}sin(wt)dt\)

\(\displaystyle u=sin(wt), \;\ du=wcos(wt)dt, \;\ v=\frac{1}{a}e^{at}, \;\ dv=e^{at}\)

\(\displaystyle \int e^{at}sin(wt)dt=\frac{e^{at}sin(wt)}{a}-\frac{w}{a}\int e^{at}cos(wt)dt\)

\(\displaystyle =\frac{e^{at}sin(wt)}{a}-\frac{w}{a}\left[\frac{e^{at}cos(wt)}{a}+\frac{w}{a}\int e^{at}sin(wt)dt\right]\)

\(\displaystyle =\frac{e^{at}sin(wt)}{a}-\frac{w^{2}}{a^{2}}\int e^{at}sin(wt)dt\)

Therefore, \(\displaystyle \left(1+\frac{w^{2}}{a^{2}}\right)\int e^{at}sin(wt)dt\)

\(\displaystyle =\int e^{at}sin(wt)dt=\boxed{\frac{e^{at}(asin(wt)-wcos(wt)}{a^{2}+w^{2}}+C}\)

Now, enter in a= -3 and you have it.

 

[Joker2k999]

Cool! yea my algebra was ok but my integration is what needed some work.

Thx guys.

JkR.

Cheers.