Integrating a complicated polynomial

[Mathpower]

Homework Statement

Hi
An integration question:
t= 1/(1+r²)

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

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?
$$\int \frac{1}{1+r^2}\,dr$$
BTW, what you have after the t is not a polynomial, but a rational expression.

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?

Yes, it's called trigonometric substitution. Try r = tan u.

 

[Mathpower]

eumyang...thank you...yes that's 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)
$= <1/(1+u) dr<br /> = <(1/(1+u)) (secu)^2 du<br /> = <((secu)^2/(1+u)) du$
Stuck...

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

 

[eumyang]

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)
$= <1/(1+u) dr<br /> = <(1/(1+u)) (secu)^2 du<br /> = <((secu)^2/(1+u)) du$
Stuck...

No no, put the two attempts together:
$$\int \frac{1}{1+r^2}\,dr = \int \frac{1}{1+\tan^2 u}\,\sec^2 u \, du$$
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
$$= ∫1 du<br /> =u + C$$
=tan(r)^-1 + C

Thank you for your help.

 

[eumyang]

oh wait is the answer u because 1+tan^2 u = sec^2 u
then <1

If you mean
$$\int 1$$
...then you're missing something.

=u
=tan(r)^-1 + C

Do you mean
$\tan r^{-1} + C$
or
$\tan^{-1} r + C$?

 

[Mathpower]

Opps... I meant $$tan^-1 r+C$$

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