# Integrating sine^x

Tags:
1. Oct 20, 2014

1. The problem statement, all variables and given/known data
Solve by variation of parameters:
y" + 3y' + 2y = sinex

2. Relevant equations
Finding the complimentary yields:
yc = c1e-x + c2e-2x

3. The attempt at a solution
I set up the Wronskians and got:
μ1 = ∫e-2xsin(ex)dx
μ2 = -∫e-xsin(ex)dx

The problem is that I have no idea how to integrate sin(ex).

I tried subbing u = e-x du = -e-x for μ1
=> ∫-u du sin(u-1)
Integration by parts either attempts to integrate sin(u-1) or endlessly integrates u without repeat.

That failed, so I tried just integrating by parts of μ1; it took 2 repetitions to get:
μ1 = -½e-2xsin(ex) - ½e-xcos(ex) + ∫½sin(ex)dx
I thought it might work if I get -ex∫½sin(ex)] in μ2
μ2 = e-xsin(ex) - ∫cos(ex)dx

Going further into the integration by parts just adds more complications, such as adding "x" as a term as well as going into higher powers of ex.

I can't express the integral as a series; that's next chapter and not covered on the mid-term in a few days (I'm currently hoping the mid-term doesn't have this problem).

2. Oct 21, 2014

### Simon Bridge

Did you try substitution: $u=e^x$.

3. Oct 21, 2014

I did initially, but found it problematic since it ends up becoming ½∫sinu/u. If you keep on doing it it winds up being a giant beast of an equation which doesn't appear to match μ2 at all.

I thought part of it might equal μ2, but that didn't happen either. The two just keep mismatching cosine and sine functions endlessly.

I just realized I did the Wronskians wrong; the bottom isn't 1... And it ends up a LOT easier, but I still can't quite seem to get it right.
With the correct Wronskian I instead got ∫sin(u)/u for μ1 and a nice and short -½cos(ex) for μ2. This does cause an elimination between the two equations, but I'm still getting a vast number of functions for μ1 that doesn't seem to have an end.

I'm going to check back on this in the morning.

4. Oct 21, 2014

### Ray Vickson

The integral $\int \sin(e^x) \, dx = \text{Si}(e^x) + C$, where $\text{Si}$ is a "non-elementary" function.

5. Oct 21, 2014

Resolved: I kept working with substituting ex = u on both μ's and it actually works insanely well.

6. Oct 21, 2014

### Simon Bridge

Well done - getting the correct Wronskian is helpful too of course ;)
(If the Wronskian was incorrect, does that make it a Wrightskian?)

Aside:
The "non-elementary function" Si(x) in post #4 is called the Sine Integral and you can look it up.

For that matter, sin(x)/x is called "sinc(x)" ... and you can look that up too.