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Integrating Exp, Trig Composite function

  • Thread starter NickKnack
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  • #1
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Homework Statement


Hey, I've been working through a book and one problem just gets me that I know should be a piece of cake. I don't know if I'm just being an idiot or not seeing something but the problem is to take int e^(ax)cos(bx)dx and int e^(ax)sin(bx)dx simultaneously by multiplying the second integral by i, combine and integrate, and then separate back out the real and imaginary parts of the answer to get both solutions. By looking on an integral table i know that my answer isn't right and i can tell that it obviously isn't right because i'm not explicitly integrating the trig functions. Anyway, here's what I have so far.


Homework Equations





The Attempt at a Solution



int(e^(ax) cos(bx))dx + i*(int e^(ax)sin(bx)dx)
int(e^(ax)(cos(bx) + isin(bx))dx)
int(e^(ax+ibx)dx)
int(e^(x(a+ib)))dx
1/(a+ib)e^(ax)(cos(bx) + isin(bx)) => 1/a*e^ax*cosbx and 1/b*e^ax*sin(bx) ?!?

I know i'm doing something simple and stupid incorrectly or maybe it's good and in the wrong form but this is driving me crazy!!!
thanks in advance for any help. oh, and sorry about the typing, i couldn't get the latex thing to work right.
 

Answers and Replies

  • #2
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I assume you intend your last expressions to be [tex]\frac1a e^{ax} \cos bx[/tex] and [tex]\frac1b e^{ax} \sin bx[/tex]; if so, you need to parenthesize: (1/a)*e^(ax)*cos(bx) and so on.

Given that: the real part of [tex]\frac1{a + ib}[/tex] is not [tex]1/a[/tex], and the imaginary part is not [tex]1/b[/tex]. Review the arithmetic of complex numbers to find how to compute the real and imaginary parts of the reciprocal of a complex number.
 
  • #3
SammyS
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To rationalize the denominator of

[tex]\displaystyle {{1}\over{a+ib}}\,,[/tex]

Multiply

[tex]\displaystyle {{1}\over{a+ib}}\cdot{{a-ib}\over{a-ib}}\,,[/tex]

without canceling factors in the numerator & denominator.
 
  • #4
2
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thanks ystael,
i figured that it was because i can't do algebra. finished it out and it works :)
 

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