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Calc II Integral

  1. Dec 10, 2011 #1
    1. The problem statement, all variables and given/known data

    http://www.wolframalpha.com/input/?i=integrate+%281%2F%28%28x^2%29sqrt%2825-%28x^2%29%29%29


    3. The attempt at a solution

    Alright, so I attempted this integral and checked my answer on wolfram and my answer was the same except for a -sin(^-1)(x/5) tagging along in mine. I can't figure out why that shouldn't be there.

    Here's what I did:

    1) Factored a 5 out of the root, so the problem then looked like:

    (1/5)∫(1/((x^2)(sqrt(1-(x/5)^2)))

    2) Set up trig substitutions:

    sec(t)=sqrt(1-(x/5)^2)
    x=5sin(t)
    dx=5cos(t)

    3) Do the substitution

    (1/5)∫(5cos(t)/(25(sin(t)^2)sec(t)))dt

    *cancel 5s, pull 25 out, replace sec(t) with 1/cos(t) yields:

    (1/25)∫(cos(t)^2)/(sin(t)^2)dt

    4) Replace cos(t)^2 with 1-sin(t)^2 and break up the fraction to get:

    (1/25)∫((csc(t)^2)-1)dt

    5) Integrate that, yielding:

    (1/25)(-cot(t) - t)

    Now, substituting the x's back in for t gives the answer on wolfram plus -(sin(x/5)^-1) which comes from solving for t using trig.

    Did I screw something up to give me the -t at the end? I know I haven't done these in a while but I can't figure out where I went wrong.

    Thanks!
     
  2. jcsd
  3. Dec 10, 2011 #2

    Tom Mattson

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    Look more closely, you didn't put the integral into Wolfram correctly. Wolfram has the radical in the denominator, whereas you have it in the numerator. Two different integrals.

    I got what you got, by the way.
     
  4. Dec 10, 2011 #3

    SammyS

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    See correction in Red.

    B.T.W.: Did you have WolframAlpha show you the steps ?
     
  5. Dec 10, 2011 #4

    Tom Mattson

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    Ack, never mind what I said. I thought the integral was supposed to be this:

    [tex]\int\frac{1}{x^2}\sqrt{1-\frac{x^2}{5^2}}dx[/tex]

    In that case you do get the arcsin term. I got a little cross-eyed with all those parentheses.
     
  6. Dec 10, 2011 #5
    bleh thanks (managed to miss that even with drawing a picture triangle). I looked at the steps in wolfram after I solved but didn't notice that our root substitutions were different.

    Thanks also Tom, and sorry about all the parentheses. I'm bad with latex. Next time I'll just scan in the sheet.
     
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