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Improper integrals and trig substitution 
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#1
Feb811, 11:55 PM

P: 16

1. The problem statement, all variables and given/known data
It's been a couple of years since I've done real math, so I'm kinda stuck on this one. This is actually part of a physics problem, not a math problem  but I'm stuck on the calculus part. I'm trying to solve this guy: [itex] \int \limits_{\infty}^{\infty} \frac{x^2}{(x^2+a^2)^2}\textrm{d}x [/itex] a is a constant 2. Relevant equations [itex]\textrm{tan}^2 \theta + 1 = \textrm{sec}^2 \theta[/itex] 3. The attempt at a solution I make the substitution [itex] x = a \textrm{tan} \theta[/itex] therefore [itex] \textrm{d}x = a\textrm{sec}^2\theta\textrm{d}\theta [/itex] giving me [itex] \int{\frac{a^2 \textrm{tan}^2 \theta a\textrm{sec}^2 \theta \textrm{d} \theta}{(a^2 \textrm{tan}^2 \theta + a^2)^2}} [/itex] and eventually I get it to boil down to (using the aforementioned tangent identity and canceling terms) [itex] \frac{1}{a} \int \textrm{tan}^2 \theta \textrm{d} \theta [/itex] I thought I was supposed to change the limits to [itex] \pm\frac{\pi}{2} [/itex] , but when I solve the above simplified integral I get [itex] \textrm{tan}\theta  \theta[/itex] which is not convergent My problem is taking the limit for the tangent at [itex]\pm\frac{\pi}{2}[/itex] I'm probably screwing up with the limits of integration. What exactly am I supposed to do with a trig substitution and the limits when dealing with an improper integral? I was following an old calculus book of mine, but this doesn't seem exactly right... Thanks for your help! 


#2
Feb911, 12:19 AM

P: 16

Just realized that it shouldn't end at
[itex]\textrm{tan}^2\theta[/itex] but [itex]\frac{1}{a} \int \frac{\textrm{tan}^2\theta}{\textrm{sec}^2\theta}\textrm{d}\theta = \frac{1}{a} \int \textrm{sin}^2 \theta \textrm{d}\theta[/itex] I'm still unsure of what exactly to do with the limits... 


#3
Feb911, 03:51 AM

Sci Advisor
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P: 11,896

I'm getting
[tex] \int_{\infty}^{\infty} \frac{x^2}{\left(x^{2}+a^{2}\right)^{2}} dx = \int_{0}^{\infty} x \, \frac{2x}{\left(x^{2}+a^{2}\right)^{2}} dx = ... [/tex] and now you can do part integration. 


#4
Feb911, 08:37 AM

P: 366

Improper integrals and trig substitution
To solve your resulting integral [itex] \int sin^2\theta \, d\theta [/itex], use the halfangle formula on [itex] sin^2 \theta [/itex]. As for the limits of integration, there are two possibilities. Once you've integrated, you can substitute back in to get expressions in x and then use the original limits. Or you can find the new limits for theta by solving
[tex] \infty = x = a \tan\theta [/tex] and [tex] \infty = x = a \tan\theta [/tex] for theta. (Hint: To do the second method, think about where cosine is 0, and whether sine is positive or negative at this point.) 


#5
Feb911, 05:46 PM

P: 16

Thanks all for your help, I think I got it!



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