integral of (secx)^3 with eulers formula

**cragar** · Jun25-09, 08:33 PM

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** · Jun25-09, 08:44 PM

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** · Jun26-09, 04:56 AM

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

**HallsofIvy** · Jun26-09, 06:25 AM

It's pretty much just algebra, isn't it?

$LaTeX Code: sec(x)= \\frac{2}{e^x+ e^{-x}}$
so
$LaTeX Code: sec^3(x)= \\frac{8}{(e^x+ e^{-x})^3}$
You can multiply both numerator and denominator by e^3x to get
$LaTeX Code: \\frac{8e^{3x}}{(e^x(e^x+ e^{-x}))^3}= \\frac{8e^{3x}}{(e^{2x}+ 1)^3}$
and your integral becomes
$LaTeX Code: \\int\\frac{8e^{3x}dx}{(e^2x+ 1)^3}$

If you let u= e^x, du= e^xdx and we have
$LaTeX Code: \\int\\frac{8u^2 du}{(u^2+ 1)^3}$
which can be done in terms of partial fractions.

**cragar** · Jun26-09, 08:15 PM

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** · Jun26-09, 08:37 PM

okay i got it now thanks

**Pengwuino** · Jun26-09, 08:40 PM

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

**snipez90** · Jun27-09, 02:21 AM

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

**cragar** · Jun27-09, 02:26 AM

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** · Jun27-09, 02:39 AM

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** · Jun27-09, 02:49 AM

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

**HallsofIvy** · Jun27-09, 06:33 AM

Sorry about dropping the "i" !

**cragar** · Jun27-09, 06:36 AM

its ok i got it now .

**cragar** · Jun28-09, 01:15 AM

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

**Count Iblis** · Jun28-09, 09:35 AM

Originally Posted by cragar

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** · Jun28-09, 10:31 AM

i see thanks