Integral of (secx)^3 with eulers formula

[cragar]

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 ?

 

[Count Iblis]

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.

 

[cragar]

can u give me an idea on how to start to integrate this.

 

[HallsofIvy]

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 e^3x 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= e^x, du= e^xdx and we have
[tex]\int\frac{8u^2 du}{(u^2+ 1)^3}[/tex]
which can be done in terms of partial fractions.

 

[cragar]

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)

 

[cragar]

okay i got it now thanks

 

[Pengwuino]

Yes it is with the imaginary exponentials. Simply replace everything with i3x and it should still follow

 

[snipez90]

Eh, seems kind of ugly. This integral has a very natural integration by parts solution.

 

[cragar]

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.

 

[snipez90]

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.

 

[cragar]

yes i agree. but i was hoping eulers formula would yield an easier soultion but appartenlty not.

 

[HallsofIvy]

Sorry about dropping the "i" !

 

[cragar]

its ok i got it now .

 

[cragar]

how do i take the arctan(e^(ix)) how do i make it into the real part.

 

[Count Iblis]

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.

 

[cragar]

i see thanks