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A tricky integral

by Appa
Tags: arctan(x), integral
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Jan14-09, 12:49 AM
P: 16
1. The problem statement, all variables and given/known data
I should find this integral:
[tex]\int[/tex]b x(1/[tex]\Pi[/tex])(1/(1+x2)dx

2. Relevant equations

[tex]\int[/tex]1/(1+x2)dx = arctan(x)

3. The attempt at a solution
The Only thing I've succeeded in doing is to take the 1/[tex]\Pi[/tex] and put it in front of the integral like this:
(1/[tex]\Pi[/tex])[tex]\int[/tex]b (x/(1+x2)dx
And I know that the integral of 1/(1+x2) equals arctan(x) but how could that help me? Ive tried to use the equation
[tex]\int[/tex]f(x)g(x)dx = F(x)g(x) - [tex]\int[/tex]F(x)g'(x)dx
but I can't compute the integral of arctan(x).
Could someone help me?
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Jan14-09, 12:54 AM
P: 21,305
Is this the integral?
[tex]\frac{1}{\pi}\int_{-b}^b{\frac{x}{x^2 + 1}dx[/tex]

If so, you can use a simple substitution, u = x^2 + 1, and du = 2xdx
Jan14-09, 12:56 AM
P: 21,305
BTW, you should have posted this in the Calculus & Beyond forum. This problem clearly falls in that area.

Jan14-09, 01:05 AM
P: 626
A tricky integral

Seems kind of weird to be doing that integral from -b to b, whether or not b > 0 or < 0 you get into complex numbers.
Gib Z
Jan14-09, 06:57 AM
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P: 3,348
Perhaps the easiest way is to observe this function is odd, and look at the interval of integration.
Jan14-09, 07:19 AM
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Quote Quote by NoMoreExams View Post
Seems kind of weird to be doing that integral from -b to b, whether or not b > 0 or < 0 you get into complex numbers.
No, you don't. You are squaring not taking square roots. As Mark44 said, use the substitution u= x2+ 1. Or, even simpler, use Gib Z's suggestion. This is really a very simple integral.
Jan14-09, 09:18 AM
P: 626
Oh duh, good point.
Jan14-09, 01:09 PM
P: 16
Yeah, sorry, I got it myself pretty soon after posting this. It feels like the more I study maths, the more I forget..!

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