anonymous
  • anonymous
Find the indefinite integral of [e^(-t)sin(t)i + e^(-t)cos(t)j] dt. side note: I'm not entirely sure of what to do with e^(-t)....
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
please use equation editor so that I can get a better look of the integration problem...thanks
anonymous
  • anonymous
\[\int\limits_{}^{} [e ^{-t}\sin(t)i + e ^{-t}\cos(t)j] dt\]
anonymous
  • anonymous
sorry, didn't know that existed haha.

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anonymous
  • anonymous
Well this is an integration by parts problem or you can use that the imaginary part of e^(i t) is sin(t) and the real part of e^(it) is cos(t).
anonymous
  • anonymous
And I assume those are i and j unit vectors?
anonymous
  • anonymous
yes they are
anonymous
  • anonymous
So how would you integrate: \[\int\limits e^{-t} \cos(t)dt\]?
anonymous
  • anonymous
\[\frac{ 1 }{ 2 }e ^{-t}(\sin(t)-\cos(t))+C \] ? i think...
anonymous
  • anonymous
Differentiate it and see if it works: \[\frac{1}{2}\frac{d}{dt} \left( e^{-t}\sin(t) -e^{-t}\cos(t) \right) = \frac{1}{2} \left( -e^{-t}\left(\sin(t) - \cos(t) \right) + e^{-t}(\cos(t) + \sin(t)\right)\] \[=\frac{1}{2}(2 e^{-t} \cos(t)) = e^{-t}\cos(t)\] This confirms your integral is correct. So then do the other one involving sine. From there you know that: \[\int\limits \left( f(x) \hat{i} + g(x) \hat{j}\right)dx = \left( \int\limits f(x) dx \right) \hat{i} + \left(\int\limits g(x) dx \right) \hat{j}\]
anonymous
  • anonymous
So simply integrate and then add on a constant vector C which encompasses the constant of integration you get from integrating the x component (the one times i hat) and the one you get from integrating the y component (the one times j hat).
anonymous
  • anonymous
\[\frac{ 1 }{ 2 }e ^{-t}(\sin(t)-\cos(t))i + ((\frac{ -1 }{ 2 }e ^{-t}(\sin(t)+\cos(t))j) + C\] ?
anonymous
  • anonymous
Yes but remember that C should be: \[\vec{C} = c_1 \hat i + c_2 \hat j\] since both the x and y components have a constant. (If you want to know how to do this in latex simply type \vec{whatever}, this will give: \[\vec{whatever}\]
anonymous
  • anonymous
I put that in, but I still got it wrong D:
anonymous
  • anonymous
on the site that I turn it in, the C is already included so I only have to type in the previous part of the answer...is there something I'm missing still?
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=integral+of+e%5E%28-t%29sin%28t%29 The x component is right http://www.wolframalpha.com/input/?i=integral+of+e%5E%28-t%29cos%28t%29 Your y component is wrong. You have a negative where it shouldn't be.
anonymous
  • anonymous
wait the -1/2 ?
anonymous
  • anonymous
nevermind, I got it. Thank you so so much!!

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