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Homework Help: Evaluate the integral (inverse trig)

  1. Mar 29, 2010 #1
    1. The problem statement, all variables and given/known data
    Evaluate the following integral:

    2. Relevant equations
    d(arcsinx)/dx = 1/sqrt(1-x2)

    3. The attempt at a solution
    I just need a good start in the right direction on this one so I can at least try it. I know the arcsin d/dx as above, but how do I make this work? Is it some kind of integration by parts or reduction? I have no idea how to go about dealing with the sqrt(2x-x2) mostly, or how it would get there from arcsin...

    any help you can give would be much appreciated!
  2. jcsd
  3. Mar 29, 2010 #2
    Try completing the square.
  4. Mar 29, 2010 #3
    hm.. I'm trying that but still stuck, I get

    int: 1/sqrt(x(2-x))dx, I can move the root up to the top easily enough by multiplying by sqrt(x(2-x))/sqrt(x(2-x)), but that doesn't help much. I found the formula relating a/sqrt(b+x2) to (a/sqrt(b))arctan(x/sqrt(b))+C, but that still doesn't help me with the 2x...
  5. Mar 29, 2010 #4
    No no. -x^2 + 2x = -(x^2 - 2x) = -(ax + b)^2 + c. Find a, b, and c.

    Once you do that, what happens if you let u = ax + b?
  6. Mar 29, 2010 #5
    by that, it should be
    -(x2-2x) = -((x-1)2-1), or a=1, b=-1, c=-1
    so then, making u=x-1, du=1, it becomes

    int{ du/sqrt(-u2-a2)

    and that falls into the arccosh formula as:
    cosh-1(1-x)+c, and that should be the final answer for the indefinate integral right? or can I just plug the -u in like that?
  7. Mar 29, 2010 #6
    Be careful. You are saying -(x-1)^2 - 1 = 2x - x^2?

    What if x = 1? Then, -1 = 2 - 1 = 1, which is wrong. It should be -(x-1)^2 + 1.
    So if we let u = x - 1
    So, we have,
    \int \frac{1}{\sqrt{1 - u^2}} du
    That should look more familiar.
  8. Mar 29, 2010 #7
    ooh, my bad... so its equal to arcsin(x-1)+C !
  9. Mar 29, 2010 #8
    Looks about right. : )
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