Integral of (secx)^3 with eulers formula

  • Thread starter cragar
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  • #1
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Main Question or Discussion Point

is it possible to integrate (secx)^3 with eulers formula
could we use that cosx = (e^(ix) + e^(-ix)) /(2)
then take it to the -3 power and multiply it out and try to integrate sec(x)^3 this way.
this is not a homework ?
 

Answers and Replies

  • #2
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This will only work well if you integrate from zero to 2 pi. But in that case this particular integral will be divergent.


A rational function of cos and sin integrated from zero to 2 pi amounts to a contour integral of a rational function over the unit circle in the complex plane, so you can directly apply the residue theorem.
 
  • #3
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can u give me an idea on how to start to integrate this.
 
  • #4
HallsofIvy
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It's pretty much just algebra, isn't it?

[tex]sec(x)= \frac{2}{e^x+ e^{-x}}[/tex]
so
[tex]sec^3(x)= \frac{8}{(e^x+ e^{-x})^3}[/tex]
You can multiply both numerator and denominator by e3x to get
[tex]\frac{8e^{3x}}{(e^x(e^x+ e^{-x}))^3}= \frac{8e^{3x}}{(e^{2x}+ 1)^3}[/tex]
and your integral becomes
[tex]\int\frac{8e^{3x}dx}{(e^2x+ 1)^3}[/tex]

If you let u= ex, du= exdx and we have
[tex]\int\frac{8u^2 du}{(u^2+ 1)^3}[/tex]
which can be done in terms of partial fractions.
 
  • #5
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thanks for doing this it must have taken you a long time ,
But when say multiply both top and bottom by e^(3x)
do you mean e^(3ix) or e(3x)
 
  • #6
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okay i got it now thanks
 
  • #7
Pengwuino
Gold Member
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Yes it is with the imaginary exponentials. Simply replace everything with i3x and it should still follow
 
  • #8
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Eh, seems kind of ugly. This integral has a very natural integration by parts solution.
 
  • #9
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i wouldn't say very natural my whole goal was to find an easier way then by parts , but i think parts is easier
but in the case of like (e^x)sinx dx this is easier with eulers formula then by parts.
 
  • #10
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Well I meant it was natural in the sense that sec^2(x) is the derivative of tan(x) and sec(x) differentiated gives sec(x)tan(x) and that really lends itself to a clean solution through integrating by parts.

As for (e^x)sinx, I would agree.
 
  • #11
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yes i agree. but i was hoping eulers formula would yield an easier soultion but appartenlty not.
 
  • #12
HallsofIvy
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Sorry about dropping the "i" !
 
  • #13
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its ok i got it now .
 
  • #14
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how do i take the arctan(e^(ix)) how do i make it into the real part.
 
  • #15
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how do i take the arctan(e^(ix)) how do i make it into the real part.
The real part of arctan[exp(ix)] is pi/4 for real x.

if f(z) is an analytic function such that for real z we have that f(z) is real, then:

f*(z) = f(z*)

The real part of f(z) is thus given by:

Re[f(z)] = [f(z) + f*(z)]/2 = [f(z) + f(z*)]/2

If we put z = exp(i x) for real x, then we have z* = 1/z, therefore:

Re[arctan(z)] = 1/2 [arctan(z) + arctan(1/z)] = 1/2 pi/2 = pi/4


The fact that

arctan(z) + arctan(1/z) = pi/2

for all z follows directly from the fact that for real z the above identity is valid using analytic continuation.
 
  • #16
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i see thanks
 

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