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Homework Help: Integrating Using Trigonometric Substitution

  1. Aug 6, 2011 #1
    1. The problem statement, all variables and given/known data

    Use the method of trigonometric substitution to evaluate the following:


    2. Relevant equations

    The only relevant equation that I could think of for this one was the trig identity:

    [itex]sin^{2}\vartheta + cos^{2}\vartheta = 1[/itex]

    3. The attempt at a solution

    This is what I got from trying this problem:


    [itex]\sqrt{4-9x^{2}}[/itex] is a difference between two squares, so it's a leg of a right triangle. So I let [itex]4sin\vartheta = \sqrt{4-9x^{2}}[/itex] to satisfy the definition of sine (Opposite Side over Hypotenuse).

    Then, I let [itex]x^{2} = \frac{4}{9}cos\vartheta[/itex] to satisfy the definition of cosine.

    Differentiating this same function with respect to [itex]x[/itex], I got:


    Substituting these into the original function:

    [itex]\int{\frac{\frac{4}{9}cos\vartheta}{4sin\vartheta} \times \frac{\frac{-2}{9}sin\vartheta}{\sqrt{\frac{4}{9}cos\vartheta}}d\vartheta}[/itex]

    [itex]\int{\frac{\frac{4}{9}cos\vartheta}{4sin\vartheta} \times \frac{\frac{-2}{9}sin\vartheta}{\frac{2}{3}\sqrt{cos\vartheta}}d\vartheta}[/itex]

    [itex]\int{\frac{1}{9} \times \frac{cos\vartheta}{sin\vartheta} \times \frac{-1}{3} \times \frac{sin\vartheta}{\sqrt{cos\vartheta}}d\vartheta}[/itex]


    At this point, I got stuck and didn't know what to do. I tried using a modified version of the identity above and just went on to see if I could get anywhere but failed.

  2. jcsd
  3. Aug 6, 2011 #2
    they hypotenuse is actually 2, remember c^2=a^2 + b^2, the other leg is actually 3x
    Last edited: Aug 6, 2011
  4. Aug 6, 2011 #3
    I can't really see how you got these substitutions... A standard substitution for this problem would be [itex]x = \frac{2}{3}\cos\vartheta[/itex]
  5. Aug 6, 2011 #4


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    It may help you to indicate that x is the variable of integration, by including dx in your integral: [itex]\displaystyle \int\frac{x^{2}}{\sqrt{4-9x^{2}}}\,dx\,.[/itex]

    To elaborate on GreenPrint's post:

    If [itex]\sqrt{4-9x^{2}}[/itex] is a leg of a right triangle, then the hypotenuse would likely be 2 and the other leg 3x.

    As you say, sine is opposite over hypotenuse, so either [itex]\displaystyle \sin(\theta)=\frac{3x}{2}[/itex] or [itex]\displaystyle \sin(\theta)=\frac{\sqrt{4-9x^{2}}}{2}\,.[/itex]

    cos(θ) is the other.
  6. Aug 7, 2011 #5
    Just making info easier to understand.

    [PLAIN]http://img856.imageshack.us/img856/4011/derpto.png [Broken]

    Make sure you find dx to get dθ.
    Last edited by a moderator: May 5, 2017
  7. Aug 7, 2011 #6


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    No, you must have [itex]x= (2/9)cos(\vartheta)[/tex].

  8. Aug 7, 2011 #7
    Oh wow. Okay. Yeah I know what I did wrong now. Thanks everyone!

    I'm still a little confused about choosing which leg to be the radical though. Does it matter? I was taught to just put the radical term on the opposite side of the reference angle. But are there instances in which this won't work?
  9. Aug 7, 2011 #8
    just think about it
    c^2 = a^+b^2

    in order to get something in the form sqrt(z^2 - y^2)
    z can only equal one thing, the hypotenuse and y can be any leg
  10. Aug 7, 2011 #9


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    No, it doesn't matter.
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