# Integral of (secx)^3 with eulers formula

## 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 ?

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.

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

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

$$sec(x)= \frac{2}{e^x+ e^{-x}}$$
so
$$sec^3(x)= \frac{8}{(e^x+ e^{-x})^3}$$
You can multiply both numerator and denominator by e3x to get
$$\frac{8e^{3x}}{(e^x(e^x+ e^{-x}))^3}= \frac{8e^{3x}}{(e^{2x}+ 1)^3}$$
$$\int\frac{8e^{3x}dx}{(e^2x+ 1)^3}$$

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

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)

okay i got it now thanks

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

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

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.

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.

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

HallsofIvy
Homework Helper
Sorry about dropping the "i" !

its ok i got it now .

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

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.

i see thanks