# a question on definite integral $\int\limits_0^t e^{\alpha t}\sin(\omega t)\,\mathrm dt$

I need to find $\int\limits_0^t e^{\alpha t}\sin(\omega t)\,\mathrm dt$. I'd like to know whether it can be brought into some closed form expression. Please suggest me some hints to solve this.

• @Carl: Is the solution $e^{\alpha t}sin(\omega t) + \frac{\omega}{\alpha^2 + \omega^2}$ correct ? Apr 23 '11 at 4:08
• I get the $\omega/(\alpha^2+\omega^2)$ part, but not the other part. Instead of it, I'm getting stuff like $\omega\cos(\omega t)/(\alpha^2+\omega^2)$ and similar for $\alpha \sin(\omega t)$, with various minus signs. Apr 23 '11 at 4:15
• My instinct is to do it as the imaginary part of the complex integral $\int \exp(\alpha t + i\omega t)\;dt$, but I see (from the Chandru1 hint) that you can do it by messing around with integration by parts. That seems messier, but will get the same answer given no errors in the algebra. Apr 23 '11 at 4:18
• @Carl: I am getting it as $e^{\alpha t} \frac{\alpha\sin(\omega t) - \omega \cos(\omega t) + \omega}{\alpha^2 + \omega^2}$ Apr 23 '11 at 4:34
• @Rajesh D; That is almost what I got, but I also have an $\exp(\alpha t)$. Note that it is zero when $t=0$, as required. When you get the right answer, you can differentiate it to see if you get the thing under the integral in the original question. Apr 23 '11 at 4:38
I would suggest to use $\sin \omega t = \frac{1}{2 i}\left(e^{i\omega t} - e^{-i\omega t}\right)$.