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Homework Help: Integrate using trig substitution

  1. Mar 27, 2008 #1
    1. The problem statement, all variables and given/known data

    int sqrt 8x-x^2

    2. Relevant equations

    trig sub

    3. The attempt at a solution

    complete the square
    integral becomes
    int sqrt 16-(x-4)^2

    let x-4= 16sin(Q) sqrt 16-(x-4)^2 =sqrt 16-256sin^2(Q)
    dx= 16cos(Q)dQ = 16cos(Q)


    int 16cos(Q)16cos(Q) dQ=256 int cos^2(Q)dQ double angle formula

    128 int 1+cos(2Q) dQ let u=2Q 1/2du=dQ

    This is the part where i get messed up...
    128 int 1 dQ + 128 int cos(u)du
    So the first part is just 128Q and the integral of 128intcos(u)du is 128sin(u)
    So do I fill 2Q back in for u? Then if I fill it back in do I use Q=[sqrt16-(x-4)^2] / 16 ??
     
  2. jcsd
  3. Mar 27, 2008 #2
    [tex]\int\sqrt{8x-x^2}dx[/tex]

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

    *[tex]x=a\sin\theta \rightarrow a=4[/tex]

    [tex]x-4=4\sin\theta[/tex]
    [tex]dx=4\cos\theta d\theta[/tex]

    [tex]\int\sqrt{4^2-4^2\cos^2\theta}d\theta[/tex]

    Re-check your substitutions!
     
  4. Mar 27, 2008 #3
    Ok so other than that does it look right? It is a definate integral so in the end i just plug my value in correct?
     
  5. Mar 27, 2008 #4
    I end up with 8[Q]+8[sin(2Q)] But im not sure how to put a value back in for Q
     
  6. Mar 27, 2008 #5
    You want it in terms of x?

    Use your substitutions ... cosine = adj/hyp and sine = opp/hyp.
     
  7. Mar 27, 2008 #6
    hmm Im not familiar with doing it like that
     
  8. Mar 27, 2008 #7
    Draw a triangle and you will know what I mean ...

    [tex]x-4=4\sin\theta[/tex]

    [tex]\sin\theta=\frac{x-4}{4}[/tex]

    [tex](x-4)^2+\cos^2\theta=4^2[/tex]

    [tex]\cos\theta=\frac{\sqrt{4^2-(x-4)^2}}{4}[/tex]
     
    Last edited: Mar 27, 2008
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