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Use trigonometric substitution to verify

  1. Nov 24, 2008 #1
    1. The problem statement, all variables and given/known data
    Using trigonometric substitution, verify that
    [tex]\int[/tex] [tex]\sqrt{a^2[-t^2}[/tex]dt (INTEGRAL FROM 0 TO [tex]\pi[/tex])=(1/2)a^2sin^{-1}(x/a)+(1/2)x[tex]\sqrt{a^2-x^2[/tex]

    Sorry it doesn't seem to want to allow me to place superscript inside a square root. but inside the first sq. root is a2-t2 and the second sq. root contains a2-x2



    2. Relevant equations

    [tex]\sqrt{a^2-x^2}[/tex]

    x=asin([tex]\theta[/tex])

    -([tex]\pi[/tex]/2) [tex]\leq[/tex] [tex]\theta[/tex] [tex]\leq[/tex] ([tex]\pi[/tex]/2)

    3. The attempt at a solution

    replacing x with asin[tex]\theta[/tex] in the original integral
    [tex]\int[/tex][tex]\sqrt{a^2-(asin^2\theta}[/tex]dt
    multiply and factor out. Change 1-sin^2[tex]\theta[/tex] to cos^2[tex]\theta[/tex]
    replace dt with dt=(-a)cos[tex]\theta[/tex] d[tex]\theta[/tex] because t=asin[tex]\theta[/tex]
    End up with
    [tex]\int[/tex] (a)(cos[tex]\theta[/tex])(-a)(cos[tex]\theta[/tex])d[tex]\theta[/tex]
    Combine and get
    -[tex]\int[/tex] a2cos2[tex]\theta[/tex] d[tex]\theta[/tex]

    At this point I considered pulling the a2 out in front since, unless I'm confused, it represents a constant and changing cos2[tex]\theta[/tex] to 1-sin2[tex]\theta[/tex] but that hasn't seemed to get me any close. Any help would be much appreciated or if I've made a mistake already identifying that for me would be excellent. Please and thank you all.
    :)
     
    Last edited by a moderator: Nov 25, 2008
  2. jcsd
  3. Nov 24, 2008 #2

    Dick

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    Science Advisor
    Homework Helper

    No, changing cos^2(theta) to sin^2(theta) doesn't help much. But using a double angle formula does. Change it to (1+cos(2*theta))/2
     
  4. Nov 25, 2008 #3
    Thanks a lot.
     
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