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Homework Help: Integration By Trig Substitution

  1. Aug 31, 2010 #1
    1. The problem statement, all variables and given/known data
    ∫√(4-x^2)/x dx


    2. Relevant equations



    3. The attempt at a solution
    a^2=4 u^2=x^2 ⇒ u=asinθ
    a=2 u=x
    x=2sinθ sinθ
    2cosθ=√(4-x^2)
    dx=2cosθ dθ

    ∫√(4-4sin^2θ)/2sinθ 2cosθ dθ
    ∫2cos^2θ/2sinθ 2cosθ dθ
    2∫cos^2θ/sinθ cosθ dθ

    Now what????????????? I have been working on this problem for the last 3 3 hours and I am at a stand still. Can somebody Please get me to the answer?????????
     
  2. jcsd
  3. Aug 31, 2010 #2

    rock.freak667

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    Now use the formula cos2θ+sin2θ=1 and replace cos2θ and then try using t=sinθ as a substitution.
     
  4. Aug 31, 2010 #3

    Dick

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    I don't think you want to use a trig substitution. Try u^2=4-x^2. Then use partial fractions on the u integral.
     
  5. Aug 31, 2010 #4
    To do partial fractions, don't I need to divide the numerator by the denominator because the exponent is larger in the numerator? If so, how? I've never tried to do synthetic division on a sqrt.
     
  6. Aug 31, 2010 #5

    Dick

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    Maybe you missed that it's u^2=4-x^2, not u=4-x^2. There's no sqrt left after the u substitution.
     
  7. Aug 31, 2010 #6
    What I have now is:
    =2∫cos^2θ/sinθ dθ
    =2∫1-sin^2θ/sinθ dθ
    =2∫1-sinθ dθ
    =2(cosθ + θ)
    But I can't resub because there is no value for θ!
     
  8. Sep 1, 2010 #7

    Dick

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    You are also making algebra mistakes. (1-sin(t)^2)/sin(t) isn't 1-sin(t). And to get things back in terms of x, you want to put theta=arcsin(x/2).
     
  9. Sep 1, 2010 #8
    well thats right

    whenever u have anything under the square root , try removing hte square root first
     
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