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Integration by substitution question

  1. Apr 22, 2003 #1
    How do you do this question, Ive spent hours figuring it out:

    Use the substitution x = 3sint to show that

    3
    [inte]x^2[squ](9-x^2) dx = (81/16)pi
    0
     
  2. jcsd
  3. Apr 22, 2003 #2
    Have you got it anywhere near [inte] 81 (sint)^2 * (cost)^2 dt from 0 to [pi]/2 yet? You can make a substitution from there. If you haven't, have you forgot to substitute dx for 3cost dt, and changed the limits?
     
  4. Apr 22, 2003 #3
    This is just a simple Trig sub problem. x=3sin(t), therefore x^2=9sin^(t), and 9-x^2=9-9sin^2(t) or 9-x^2=9cos^2(t), and dx=3cos(t)so the problem becomes:

    3
    [inte] 9sin^2(t)*3cos(t)*3cos(t)dt
    0

    the factor out the constants and then sub sin^2(t) as (1-cos^2(t)), then distribute the other cos^2(t), and bust out an integration table for cos^2(t) and cos^4(t).. that's about all I can tell ya without actually performing the written instructions. Hope this helps in future endeavors as well as the current problem. :smile:
     
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