Integration via Trigonometric Substitution

  • #1
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Homework Statement


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.

Homework Equations




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:

Answers and Replies

  • #2
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Try to change the last integral by using other trigonometric functions, like ## \tan\theta ## and ## \sec\theta.##
 
  • #3
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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.
 
  • #4
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So what is wrong with this:
[tex]\int{\frac{sin^{2}θ}{(cos^{2}θ)^\frac{5}{2}}}dθ=
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.

Yeah I forgot to type that in, anyway it's trig identity I'm kinda having trouble with atm.
 
  • #5
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## dx ## cannot change directly ## d\theta ##. They have to connected by some function.
 
  • #6
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## dx ## cannot change directly ## d\theta ##. They have to connected by some function.

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]
 
  • #7
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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
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Then, what is ## dx ## as a function of ## \theta ##?
 
  • #10
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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.
 
  • #11
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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.
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.
 
  • #12
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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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