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Antiderivative of square root function

  1. Mar 11, 2008 #1
    1. find the antiderivative of sqrt(81-x^2) in the bounds 0 to 9/2

    okay so the first thing i did was sin substitution.
    where I made x = 9sin(t) dx = 9cos(t) where t is theta.
    then after some manipulating i got that that new integral was 9cos^2(t)
    i then used the double angle formula and made it equal to 9/2 * (1 + cos(2t))
    I then found the antiderivative of this integral to be 9/2 * (t + sin(2t)/2)
    Using the identity that sin(2t) = 2sin(t)cos(t) the antiderivative became
    (9/2) (t + cos(t)sin(t))
    I then changed the bounds of the integral.
    When x goes from 0 to 9/2, theta goes from 0 to pi/6
    i then solved the integral to get 4.304
    The actual answer is 38.74 so i think im completely missing the boat here.
  2. jcsd
  3. Mar 11, 2008 #2
    Ps any tips you might have would really help me out it doesnt have to be too in detail im usually pretty good at this stuff. Thank you!!!!!!!
  4. Mar 11, 2008 #3
    okay so it was actually a really dumb sneaky mistake:
    after all the subtituting and manipulating was done instead of 9cos^2(t) its actually supposed to be 81cos^2(t) then follow my steps from there and its right!!
  5. Mar 11, 2008 #4
    It is a standard integral [i.e. anti-derivatives of a general form of problems]:

    \int \sqrt{a^2 - x^2}dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} sin^{-1}\left(\frac{x}{a}\right) + C
  6. Mar 11, 2008 #5
    thats really cool! ive never seen that formula before!!
  7. Mar 12, 2008 #6


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    Science Advisor

    It is, honestly, better to know HOW to do the integral, the way you did, rather than remember (or even look up) a formula.
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