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January 31st, 2018, 08:24 PM   #1
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Evaluate an Integral

∫e^(-x)cos(x)dx

Last edited by skipjack; February 8th, 2018 at 09:50 AM.
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January 31st, 2018, 08:27 PM   #2
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Quote:
Originally Posted by margaretk View Post
∫e^(-x)cos(x)dx
If you aren't going to bother attempting the exercise for yourself, or explain any difficulties you are having, I suggest you pay for a Wolfram Alpha subscription and get worked solutions there.

Wolfram|Alpha: Computational Knowledge Engine

Last edited by skipjack; February 8th, 2018 at 09:51 AM.
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January 31st, 2018, 08:36 PM   #3
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Sure, sure, I know the usual way, but I would actually prefer to work out
$$\int e^{-x} e^{ix}dx$$
instead. The solution follows from this, and you get one for free too.
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February 8th, 2018, 04:52 AM   #4
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Specifically, what microm@ss is suggesting is that, since $\displaystyle \cos(x)= \frac{e^{ix}+ e^{-ix}}{2}$, $\displaystyle e^x \cos(x)= e^x\left(\frac{e^{ix}+ e^{-ix}}{2}\right)= \frac{1}{2}\left(e^{(1+ i)x}+ e^{1- i)x}\right)$.

Can you integrate that?

Another way, using integration by parts:
Let $\displaystyle u= \cos(x)$, $\displaystyle dv= e^x dx$. Then $\displaystyle du= -\sin(x) dx$ and $\displaystyle v= e^x$. So, by "integration by parts" ($\displaystyle \int u dv= uv- \int vdu$)
$\displaystyle \int e^x \cos(x)= e^x \cos(x)- \int (-\sin(x))e^x dx= e^x \cos(x)+ \int e^x \sin(x)dx$.

To do that integral, use integration by parts again. Let $\displaystyle u= \sin(x)$ $\displaystyle dv= e^x dx$. Then $\displaystyle du= \cos(x)dx$ and $\displaystyle v= e^x$ so that $\displaystyle \int e^x \sin(x)= e^x \sin(x)- \int e^x \cos(x) dx$.

Putting that into the previous integral,
$\displaystyle \int e^x \cos(x) dx= e^x \cos(x)+ e^x \sin(x)- \int e^x \cos(x) dx$.

Now, add $\displaystyle \int e^x \cos(x) dx$ to both sides to get
$\displaystyle 2\int e^x \cos(x) dx= e^x \cos(x)+ e^x \sin(x)$

and then
$\displaystyle \int e^x \cos(x) dx= \frac{1}{2}e^x(\cos(x)+ \sin(x))+ C$

Last edited by skipjack; February 8th, 2018 at 09:54 AM.
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February 10th, 2018, 08:47 PM   #5
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For those not familiar with that fact, using integration by parts combined with a trig substitution may be useful here.
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