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Integration via Trigonometric Substitution

  1. Oct 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Evaluate [tex]\int{\frac{x^2}{(1-x^2)^\frac{5}{2}}}dx[/tex] via trigonometric substitution.
    You can do this via normal u-substitution but I'm unsure of how to evaluate via trigonometric substitution.
    2. Relevant equations


    3. The attempt at a solution
    Letting [tex]x=sinθ[/tex],
    [tex]\int{\frac{sin^{2}θ}{(1-sin^{2}θ)^\frac{5}{2}}}dθ=\int{\frac{sin^{2}θ}{(cos^{2}θ)^\frac{5}{2}}}dθ[/tex]
    but I'm not sure how the working in the answers gets up to [tex]\int{\frac{x^2}{(1-x^2)^\frac{5}{2}}}dx=\int{\frac{sin^{2}θ}{cos^{4}θ}}dθ[/tex].
     
    Last edited: Oct 10, 2015
  2. jcsd
  3. Oct 10, 2015 #2
    Try to change the last integral by using other trigonometric functions, like ## \tan\theta ## and ## \sec\theta.##
     
  4. Oct 10, 2015 #3
    Oh, I misunderstand what you want to know. I thought you want to calculate the last thing.

    When you starts from the beginning, there should be ## dx##, the integral variable. Then, I think you may find what you missed during your calculation.
     
  5. Oct 10, 2015 #4
    So what is wrong with this:
    [tex]\int{\frac{sin^{2}θ}{(cos^{2}θ)^\frac{5}{2}}}dθ=
    Yeah I forgot to type that in, anyway it's trig identity I'm kinda having trouble with atm.
     
  6. Oct 10, 2015 #5
    ## dx ## cannot change directly ## d\theta ##. They have to connected by some function.
     
  7. Oct 10, 2015 #6
    Not too sure what you mean by that.
    For functions with the form [tex]\sqrt{a^2-x^2}[/tex] you can express them as [tex]x=asinθ[/tex]
     
  8. Oct 10, 2015 #7
    See,

    [tex] \int \dfrac{x^2}{(1-x^2)^{5/2}} dx = \int \dfrac{ \sin^2\theta}{(1-\sin^2\theta)^{5/2}} dx = \int \dfrac{\sin^2 \theta}{\cos^5\theta} dx \neq \int \dfrac{\sin^2 \theta}{\cos^5\theta} d\theta.[/tex]
    You missed something in the lase step.
     
  9. Oct 10, 2015 #8
    Yeah it was a typo...
     
  10. Oct 10, 2015 #9
    Then, what is ## dx ## as a function of ## \theta ##?
     
  11. Oct 10, 2015 #10
    The denominator is somehow supposed to be [tex]cos^{4}θ[/tex], not [tex]cos^{5}θ[/tex]. That's all I need help with, nothing else.
     
  12. Oct 10, 2015 #11
    Yes, I know and you almost reach the final goal.

    ##dx## have to change as ## d\theta ## because the last integration is in a form ## \int (\cdots) d\theta ##. But as you know, ## dx \neq d\theta ##.

    What is ## dx / d\theta ##? It's really strong hint about this problem.
     
  13. Oct 10, 2015 #12
    I hope you already got the answer for the problem.

    The key idea is chain rule. The differentiation of ## x## with respect to ## \theta ## is ## dx/d\theta = \cos\theta ##. Then, the integration will change as
    $$ \int \dfrac{\sin^2\theta}{\cos^5\theta} dx = \int \dfrac{\sin^2\theta}{\cos^5\theta} \dfrac{dx}{d\theta} d\theta = \int \dfrac{\sin^2\theta}{\cos^5\theta} \cos \theta d\theta = \int \dfrac{\sin^2\theta}{\cos^4\theta} d\theta.$$
     
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