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Homework Help: Trigonometric Substitution Problem

  1. Jun 3, 2010 #1
    This problem looks relatively simple, but the coefficient in front of the variable is causing issues:

    [tex]\int{\sqrt{1-4x^{2}}}dx[/tex]

    So I started like this:

    [tex]x=sin(\theta)[/tex]
    [tex]dx=cos(\theta)d\theta[/tex]
    [tex]\int{\sqrt{1-4sin^{2}(\theta)}cos(\theta)d\theta}[/tex]

    Normally you can remove the constant from the root and go about the rest of the integral, but I'm stumped on this. How can I break this down into something more manageable? I've tried several identities but nothing has worked yet.

    A pointer in the right direction would be great. Thanks in advance!
     
  2. jcsd
  3. Jun 3, 2010 #2

    Mark44

    Staff: Mentor

    Use the substitution sin(theta) = 2x. Rather than trying to remember which substitution to use, I draw a right triangle, labellings its sides according to the quantities in the radical. My triangle has a hypotenuse of 1, opposite side of 2x, and adjacent side of sqrt(1 - 4x^2).
     
  4. Jun 3, 2010 #3
    So you just treat the root as the Pythagorean theorem? Makes sense.

    To make sure I understand this fully, if my equation was [tex]\int{\sqrt{1+16x^{2}}}dx[/tex]

    I would say that one of triangle's sides was 1, and the other was 4x? Making the hypotenuse the square root of 1-16x^2?

    Okay, so going back to the original I can say that sin(theta) = 2x/1, and then isolate x to get x=sin(theta)/2. Putting that in it squares and cancels out the 4. Wonderful, thanks for the help!
     
  5. Jun 3, 2010 #4

    Mark44

    Staff: Mentor

    Yes.
    No, the hypotenuse would be sqrt(1 + 16x^2).
     
  6. Jun 3, 2010 #5
    I hit the minus sign by accident. :p
     
  7. Jun 3, 2010 #6

    Mark44

    Staff: Mentor

    I hope you don't get a job at the bank I go to :yuck:

    (Just kidding!)
     
  8. Jun 4, 2010 #7
    Eww...banking.:yuck:
     
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