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Integral, hmm where did i mess up?

  1. Nov 10, 2007 #1
    [SOLVED] Integral, hmm ... where did i mess up?

    ok, i typed this integral into mathematica, my last part is wrong or perhaps it's equivalent, but not obvious to me

    thanks for the help!

    in mathematic ... (x^2 -1)/(x^2 + 1)^3

    Last edited: Nov 10, 2007
  2. jcsd
  3. Nov 10, 2007 #2
    I believe where you have [tex] \int^{1}_{0} \frac{tan^2 \theta}{sec^4 \theta} - \frac{1}{sec^4 \theta} [/tex], you should have [tex]sec^5 \theta [/tex], instead of to the 4th power.
  4. Nov 10, 2007 #3

    D H

    Staff: Mentor

    You forgot to transform the integration limits when you switch from integrating wrt x to integrating wrt theta.
  5. Nov 10, 2007 #4
    if i have


    isn't that the power rule? so

  6. Nov 10, 2007 #5
    Oops, yes you're correct. But it still wouldn't be to the 4th power.
  7. Nov 10, 2007 #6
    why not? and what should it be?
  8. Nov 10, 2007 #7
    But it looks like he changed it back to "x," so it wouldn't matter right?
  9. Nov 10, 2007 #8
    Well if [tex] x = tan \theta [/tex], and [tex] tan^2 \theta + 1 = sec^2 \theta [/tex], then you have [tex] (sec^2 \theta)^3 [/tex] in the denominator.
  10. Nov 10, 2007 #9
    i simplified

  11. Nov 10, 2007 #10
    OK ok, I see, let me go back and re-work it.
  12. Nov 10, 2007 #11
    thanks a lot :-] i just don't know wth is wrong with that last part, lol.
  13. Nov 10, 2007 #12

    D H

    Staff: Mentor

    Roco, you did the trig part right. The problem is the integration limits.


    He made the same mistake twice, yet another example of two wrongs making a right. I answered this question,

    Upon switching the integration variable from [itex]x[/itex] to [itex]\theta[/itex], roco, you should have switch the integration limits as well:

    [tex]\int_0^1 \frac{x^2-1}{(x^2+1)^3}dx[/tex]

    [tex]\int_{\tan^{-1} 0}^{\tan^{-1} 1} 1+2\cos2\theta + \cos4\theta d\theta[/tex]

    Evaluating the limits, [itex]\tan^{-1} 0=0[/itex] but [itex]\tan^{-1} 1=\pi/4[/itex], not 1.

    BTW, this integrates directly to

    [tex]-\;\frac 1 4 \left.\left(\theta + \sin2\theta+\frac 1 4\sin4\theta\right)\right|_0^{\pi/4}[/tex]
    Last edited: Nov 10, 2007
  14. Nov 10, 2007 #13
    why would they change? i re-substituted or whatever you call it. hmm
  15. Nov 10, 2007 #14

    D H

    Staff: Mentor

    I updated my IN WORK post. Do you see now why you have to pay attention to the integration limits? When you do so, you get the correct results.
  16. Nov 10, 2007 #15
    but that's not what i'm trying to do. when using trig subst. i don't HAVE TO switch limits do i?

    i fix my limits for other problems, but i'm just trying to work this one all the way.
  17. Nov 10, 2007 #16
  18. Nov 11, 2007 #17

    can someone link me to a site describing well the transformation of integral limits, ive been shown alot but never understood it well, thanks.

    anyway i got up to [tex] \int \frac {d \theta}{sec^4 \theta}[/tex]
    i dont know what limits to put in there
  19. Nov 11, 2007 #18
    all you do is evaluate your current limits with your substitution

    [tex]\int_0^1 \frac{x^2-1}{(x^2+1)^3}dx[/tex]

    my limits are from 0 to 1

    my chosen substitution is tangent, so i set my current limits equal to my substitution

    tangent = 0 and tangent = 1

    where does tangent equal 0 and 1? solve for that, and those are your new limits, respectively



    this is done so that there is no need to re-substitute.
  20. Nov 11, 2007 #19

    D H

    Staff: Mentor

    Suppose you can rewrite some function [itex]f(x)[/itex] as [itex]h(g(x))g'(x)[/itex]. This is what makes u-substitution viable for solving a integration problem. Applying the u-substitution [itex]u=g(x)[/itex] to the indefinite integral of f(x),

    \int f(x) dx = \int h(g(x))g'(x)dx
    \begin{matrix} \\[-8pt] \longrightarrow \\[-10pt] ^{u=g(x)} \end{matrix}
    \int h(u) du

    In applying the u-substitution to a definite integral, you must apply the substitution to the integration limits as well as to the integrand:

    \int_{x=a}^{x=b} f(x) dx = \int_{x=a}^{x=b} h(g(x))g'(x)dx
    \begin{matrix} \\[-8pt] \longrightarrow \\[-10pt] ^{u=g(x)} \end{matrix}
    \int_{u=g(a)}^{u=g(b)} h(u) du

    In this case, the u-substitution is [itex]x=\tan\theta[/itex] or [itex]u=\tan^{-1}x[/itex]. Applying the u-substitution to this problem,

    \int_{x=0}^{x=1} \frac{x^2-1}{(x^2+1)^3}dx\,
    \begin{matrix} \\[-8pt] \longrightarrow \\[-8pt] ^{\theta=\tan^{-1}(x)} \end{matrix}\,
    \int_{\theta=\tan^{-1} (0)}^{\theta=\tan^{-1} (1)} 1+2\cos2\theta + \cos4\theta d\theta

    Finally [itex]\tan^{-1}(0) = 0[/itex] and [itex]\tan^{-1}(1) = \pi/4[/itex].
  21. Nov 11, 2007 #20

    D H

    Staff: Mentor

    Spam! Somebody please delete the previous post (#20). As that will make this post #20, you can delete it, too.
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