# Need help with an integration problem

• June 7th 2011, 01:13 PM
Lancet
Need help with an integration problem
Here's what I'm working with:

$\int e^xcos(x)dx$

This does not appear to be a trivial problem, and I'm having trouble figuring out how to approach this. If I try integrating it by parts, I just end up with another integral that's very similar. Tossing this into WolframAlpha shows a *really* complicated u-sub. I've run into situations where WolframAlpha makes things a lot more complicated than they need to be. I'm hoping this is one of those times. :)

Can anyone give me a hand?
• June 7th 2011, 01:17 PM
TheCoffeeMachine
Define it with its corresponding integral and integrate both by parts:

\begin{aligned} & I : = \int e^x\cos{x}\;{dx}, ~~ J := \int e^x \sin{x}\;{dx} \\& I = \int e^x \left(\sin{x}\right)'\;{dx} = e^x\sin{x}-\int e^x\sin{x}\;{dx} = e^x\sin{x}-J. \\& J = -\int e^x \left(\cos{x}\right)'\;{dx} = -e^x\cos{x}+\int e^x\cos{x}\;{dx} = -e^x\cos{x}+I. \end{aligned}

Now by abstracting these two integrals from each other, we have:

\begin{aligned} & I-J = e^x\sin{x}-J+e^x\cos{x}-I \\& \Rightarrow 2I = e^x\sin{x}+e^x\cos{x} \\& \Rightarrow I = \frac{1}{2}e^x \left(\sin{x}+\cos{x}\right). \end{aligned}
• June 7th 2011, 08:37 PM
Lancet
Definitely not a technique I'm familiar with. I'm going to have to stare at this for a while in order to comprehend it fully. Thank you for the insight!
• June 7th 2011, 09:25 PM
chisigma
Quote:

Originally Posted by Lancet
Here's what I'm working with:

$\int e^xcos(x)dx$

This does not appear to be a trivial problem, and I'm having trouble figuring out how to approach this. If I try integrating it by parts, I just end up with another integral that's very similar. Tossing this into WolframAlpha shows a *really* complicated u-sub. I've run into situations where WolframAlpha makes things a lot more complicated than they need to be. I'm hoping this is one of those times. :)

Can anyone give me a hand?

A 'suggestive' solution is based on the 'identity'...

$e^{x}\ \cos x = \frac{e^{(1+i) x} + e^{(1-i) x}}{2}$ (1)

... and You can integrate both terms of (1) using the standard rule...

$\int e^{a x} dx = \frac{e^{a x}}{a} + c$ (2)

Kind regards

$\chi$ $\sigma$
• June 7th 2011, 11:59 PM
tom@ballooncalculus
Quote:

Originally Posted by Lancet
Definitely not a technique I'm familiar with. I'm going to have to stare at this for a while in order to comprehend it fully. Thank you for the insight!

Just in case a picture helps (helps me by mapping out the relationships implied by combinations of relative pronouns, integral of, derivative of) ...

http://www.ballooncalculus.org/asy/maps/parts.png

Key:
Spoiler:
http://www.ballooncalculus.org/asy/prod.png

... is the product rule: straight continuous lines differentiate downwards with respect to x.

... as lazy integration by parts, doing without u and v and du and dv (as does THC, too, to an extent) ... then you want 'integration by parts twice' i.e. filling the blanks here...

http://www.ballooncalculus.org/draw/parts/twice/two.png

... and then solving the top row for I.

Larger

__________________________________________________ __________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

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• June 10th 2011, 10:58 AM
Lancet
I believe I understand it now. Thanks for helping, everyone!