Integrating a complicated polynomial

Mathpower

1. The problem statement, all variables and given/known data
Hi
An integration question:
t= 1/(1+r²)

Can you please show me how to integrate with respect to r. Thank you.



3. The attempt at a solution
∴t = (1+r²)^-1
I then tried substituting u = 1+r^2 but that didn't work!
Is there a trick with this?

Vorde

Well I'll hint and say it has to do with an inverse trig function.

eumyang

Just so I understand, you're trying to find the integral of this?
[tex]\int \frac{1}{1+r^2}\,dr[/tex]
BTW, what you have after the t is not a polynomial, but a rational expression.
Yes, it's called trigonometric substitution. Try r = tan u.

Mathpower

eumyang...thank you...yes thats what i meant.

Mathpower

NO idea! I tried r= tan u
1/(1+(tan u)^2)...
Too me that just seems even more worse!

Then (LET: < be integral sign/ integrand)
[latex]
= <1/(1+u) dr
= <(1/(1+u)) (secu)^2 du
= <((secu)^2/(1+u)) du [/latex]
Stuck...

and my latex is not working as well? Have I typed the wrong code?

eumyang

No no, put the two attempts together:
[tex]\int \frac{1}{1+r^2}\,dr = \int \frac{1}{1+\tan^2 u}\,\sec^2 u \, du[/tex]
You know your trig identities, right?

EDIT: Regarding LaTeX, that's because you're using the wrong tags. It should be tex or itex in the brackets, not latex.

Mathpower

Delete this. See below

Mathpower

Oh wait is the answer:
u because 1+tan^2 u = sec^2 u
then
[tex] = ∫1 du
=u + C [/tex]
=tan(r)^-1 + C

Thank you for your help.

eumyang

If you mean
[tex]\int 1[/tex]
...then you're missing something.
Do you mean
[itex]\tan r^{-1} + C[/itex]
or
[itex]\tan^{-1} r + C[/itex]?

Mathpower

Opps... I meant [tex] tan^-1 r+C [/tex]

Thank you so much for your help. Sorry for being lazy.