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A tricky integral 
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#1
Jan1409, 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+x^{2})dx ^{b} 2. Relevant equations [tex]\int[/tex]1/(1+x^{2})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+x^{2})dx ^{b} And I know that the integral of 1/(1+x^{2}) 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? 


#2
Jan1409, 12:54 AM

Mentor
P: 21,216

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 


#3
Jan1409, 12:56 AM

Mentor
P: 21,216

BTW, you should have posted this in the Calculus & Beyond forum. This problem clearly falls in that area.



#4
Jan1409, 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.



#5
Jan1409, 06:57 AM

HW Helper
P: 3,352

Perhaps the easiest way is to observe this function is odd, and look at the interval of integration.



#6
Jan1409, 07:19 AM

Math
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Thanks
PF Gold
P: 39,327




#7
Jan1409, 09:18 AM

P: 626

Oh duh, good point.



#8
Jan1409, 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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