Trig substitution integral (I hope)

[ArcanaNoir]

Homework Statement

I got to a place in a problem where I need to do a sticky integral, and I'm hoping I can use a trig substitution. If not, I will need to solve the main problem another way :(

[tex] \int_0^\infty \sqrt{1+(e^{-\theta })^2} \; \mathrm{d} \theta [/tex]

Homework Equations

[tex] 1+\tan ^2 \theta =\sec ^2 \theta [/tex]

The Attempt at a Solution

Can I let [itex] e^{- \theta } = \tan \phi [/itex] ?
if so, does [itex] \mathrm{d} \theta = \sec ^2 \phi \; \mathrm{d} \phi [/itex] ?

And then, do I have
[tex] \int \sec ^3 \phi \; \mathrm{d} \phi [/tex] ?

 

[Dick]

Aside from the fact you can tell the integral is divergent just by looking at it, d(exp(-t))=(-exp(-t)*dt).

 

[I like Serena]

Any particular reason you want to do a trig substitution?

Myself, I'm not a fan of trig substitutions.
Usually my first step is take a part of the expression and call it "u".
Do a substitution and see what you are left with.
I see a trig substitution more as a last resort.

 

[ArcanaNoir]

Maybe I should start at the beginning.

I need to find the total length of the spiral [itex] r(\theta )=e^{-\theta } [/itex] for [itex] \theta \in [0,\infty ) [/itex]

There is a formula for arc length, but I don't necessarily have to use it if there is another way:

[tex] L=\int \sqrt{1+(r')^2} \; \mathrm{d} \theta [/tex]

I'm concerned right away about the infinity.

 

[I like Serena]

Hmm, that is not the right formula for the length of your curve.
I suspect you are mixing up the formula with a cartesian version.

 

[ArcanaNoir]

Alas. So, how about [tex] L=\int_0^\infty \sqrt{r^2+(\frac{dr}{d\theta} )^2} \; \mathrm{d} \theta [/tex]

in which case I will be computing
[tex] \int_0^\infty \sqrt{ e^{-2 \theta } + (-e^{-\theta } )^2 } \; \mathrm{d} \theta [/tex] ?

 

[I like Serena]

That looks much better! ;)

 

[ArcanaNoir]

Okay, thanks. night-night!