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Integral of {sqrt(4-x^2)} from 0 to 2?

  1. Apr 23, 2008 #1
    [SOLVED] Integral of {sqrt(4-x^2)} from 0 to 2?

    Can someone please explain step by step for this?
    I know the answer is pi, I just don't understand the steps towards solving it

    First, I'm supposed to set x = 2sin(u)
    so then u = arcsin(x/2)
    then somehow, {sqrt(4-x^2)} = 2cos(u)
    How'd that happen? @__@
    I'm really really confused!

    Thanks for tolerating XD
  2. jcsd
  3. Apr 23, 2008 #2


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    Okay so we have,

    [tex]\int^{2}_{0} \sqrt{4-x^2}\;\;dx[/tex]

    And if we let [itex]x = 2\sin u[/itex], then we obtain,

    [tex]\int^{2}_{0} \sqrt{4-\left(2\sin u\right)^2}\;\;dx[/tex]

    [tex] = \int^{2}_{0} \sqrt{4-4\sin^2 u}\;\;dx[/tex]

    [tex] = \int^{2}_{0} \sqrt{4\left(1-\sin^2 u\right)}\;\;dx[/tex]

    [tex] = \int^{2}_{0} 2\sqrt{1-\sin^2 u}\;\;dx[/tex]

    However, we know that [itex]\sin^2\theta+\cos^2\theta = 1 \Rightarrow 1-\sin^2\theta = \cos^2\theta[/itex]. Hence, the integral becomes,

    [tex]\int^{2}_{0} 2\cos u \;\;dx[/tex]

    Is that clear? Don't forget that before you can integrate, you need to make a change of variable from dx to du.

    P.S. We have Homework & Coursework forums for textbook questions and assistance with homework and we have Mathematics forums for general mathematics discussions.
  4. Apr 23, 2008 #3
    Ah! Yes! It helps very much!
    Thank you!!
    And sorry for the misplaced topic >__<;;
  5. Apr 23, 2008 #4


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    A pleasure. And don't worry about misplacing your topic, we'll let you off... this time :devil:
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