How to integrate this?

1. Oct 30, 2014

Math10

1. The problem statement, all variables and given/known data
Integrate e^(ax+by)(a*cos(xy)-y*sin(xy)) with respect to x.

2. Relevant equations
None.

3. The attempt at a solution
The answer is (cos(xy))(e^(ax+by)). I know that I need to treat y as a constant since I'm integrating it with respect to x but I don't know how to integrate this at all. Please help me.

2. Oct 30, 2014

Zondrina

Here is the equation in a LaTeX format:

$$\int e^{ax+by}(a*cos(xy) - ysin(xy)) dx = a \int e^{ax+by}cos(xy) dx - y \int e^{ax+by}sin(xy) dx$$

I believe integration by parts should take it from here.

3. Oct 30, 2014

Math10

Let me try.

4. Oct 30, 2014

Math10

I tried integration by parts but it seems that it complicates the problem even more.
Here's my work:
u=cos xy
dv=ae^(ax+by) dx
du=-y*sin(xy)
v=e^(ax+by)
(cos(xy))(e^(ax+by))-integral of (e^(ax+by))(-y*sin(xy))

5. Oct 30, 2014

Zondrina

Have you been introduced to the table method? It makes integration by parts a little bit easier to manage.

I hope you have a lot of paper laying around for this one.

6. Oct 30, 2014

RUber

Have you seen the exponential substitution for $\sin$ and $\cos$?
$\sin xy = \frac{e^{ixy}-e^{-ixy}}{2i}, \quad \cos xy = \frac{e^{ixy}+e^{-ixy}}{2}$.
Handling this integration on exponentials will be much simpler.

7. Oct 30, 2014

RUber

Nevermind, this looks like a better candidate for substitution.
Try $u = e^{ax+by} a cos(xy)$. What would du be?

8. Oct 30, 2014

Staff: Mentor

And the above can be simplified slightly by bringing eby outside each integral. The property I'm using is that eu + v = eu * ev. Since the integration is to be done with respect to x, any factors involving just y can be considered constants, and brought out of the integration.

To do each integral, you'll need to use integration by parts twice. It will seem that you're going around in circles, as you end up with something similar to what you started with (if you do it right), but you can solve algebraically for the integral. Your textbook probably has an example of this technique.

9. Oct 31, 2014

vela

Staff Emeritus
After you pull the $e^{by}$ out as Mark suggested, you're left with an integrand of $ae^{ax}\cos xy - e^{ax}y\sin xy$. My first thought upon seeing that integrand was that it looked like it might be the result of the product rule because $ae^{ax}$ is the derivative of $e^{ax}$. A quick check verified that it was indeed, making the integration trivial. Doing stuff like that can help you avoid a lot of unnecessary work.

10. Oct 31, 2014

Math10

How does the table method of integration by parts work? I've never been taught using that method.