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Homework Help: Integral of √(a^2 - x^2)

  1. Oct 5, 2011 #1
    1. The problem statement, all variables and given/known data
    ∫√(a2 - x2) dx

    2. Relevant equations

    3. The attempt at a solution
    What I've understood so far is its a substitution integral.
    Through the instruction of my lecture notes, I have

    ∫√(a2 - x2) dx Let u = sin-1(x/a)
    x = a*sin(u)
    dx = a*cos(u)
    This means that √(a2 - x2) = √(a2 - a2sin2(x))

    Then as the trig identity of √(cos2(u) + sin2(u)) = 1.
    You can rearrange it to give cos(u) = √(1-sin2(u))
    So √(a2 - a2sin2(u)) = √(a2 * cos2(u)) = a*cos(u)

    Now there is my problem, A lot of other sources say that rather than ∫a*cos(u) du
    it is ∫a2 * cos2(u) du
    and I don't understand why.
  2. jcsd
  3. Oct 5, 2011 #2


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    OK, so you rewrote the radical in terms of u, so at this point you have
    [tex]\int \sqrt{a^2-x^2}\,dx = \int a\cos u\,dx[/tex]Note that the integral still has dx in it, not du.
  4. Oct 5, 2011 #3
    So it is ∫a*cos(u) dx and not ∫a2 * cos2 (u) dx?
    I'm just struggling as I don't understand why some people are using ∫a2 * cos2 (u) dx.
    I've checked on wolfram Alpha also, and there is also the ∫a2 * cos2 (u) dx and I can't reconcile this from my above working.
  5. Oct 5, 2011 #4
    You still have the dx in there. From your first post, what does dx =? Put it in terms of du.
  6. Oct 5, 2011 #5


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    Yes, but you don't want it to be that- you want everything in terms of u, not x.

    NO, it isn't! Look again! You cannot change everything except the differential to u but leave the differential as "dx".
    With the substitution x= asin(u), dx= d(asin(u))/du du= acos(u)du.

    Now, you have both [itex]\sqrt{a^2- x^2}= \sqrt{a^2- a^2sin^2(u)}= \sqrt{a^2cos^2(u)}= a cos(u)[/itex] (as long as we are careful about signs) and [itex]dx= a cos(u)du[/itex]. Putting those together, [itex]\int\sqrt{a^2- x^2} dx= \int (a cos(u))(a cos(u)du)= a^2\int cos^2(u)du[/itex].
  7. Oct 6, 2011 #6
    Oh of course!
    I can't believe I didn't see that.
    Thank you very much!
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