# Integrating a complicated polynomial

1. Aug 11, 2012

### 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?

2. Aug 11, 2012

### Vorde

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

3. Aug 11, 2012

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

4. Aug 12, 2012

### Mathpower

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

5. Aug 12, 2012

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

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

Last edited: Aug 12, 2012
6. Aug 12, 2012

### eumyang

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.

7. Aug 12, 2012

### Mathpower

Delete this. See below

Last edited: Aug 12, 2012
8. Aug 12, 2012

### Mathpower

u because 1+tan^2 u = sec^2 u
then
$$= ∫1 du =u + C$$
=tan(r)^-1 + C

Last edited: Aug 12, 2012
9. Aug 12, 2012

### eumyang

If you mean
$$\int 1$$
...then you're missing something.
Do you mean
$\tan r^{-1} + C$
or
$\tan^{-1} r + C$?

10. Aug 12, 2012

### Mathpower

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

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