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Changing form of integral

  1. Jun 11, 2013 #1
    1. The problem statement, all variables and given/known data

    Use integration by parts to show that

    ∫(sqrt(1-x^2) dx = x(sqrt(1-x^2) + ∫ (x^2) / sqrt(1 - x^2)

    write x^2 = x^2 -1 + 1 in the second integral and deduce the formula


    ∫(sqrt(1-x^2) dx = (1/2)x(sqrt(1-x^2) + (1/2)∫ 1 / sqrt(1 - x^2) dx

    I actually found a solution guide to this problem online. Although I don't even understand the solution. I do but I'm stuck on something. Please see the pdf attached. It is exercise 3.

    2. Relevant equations



    3. The attempt at a solution


    OK so I'm confused on how they got the first integral after the = sign which reads -∫sqrt(1-x^2) dx

    So what I tried to do was what the book tells me to do.

    I sub so I got ∫x^2 / sqrt(1-x^2)dx = ∫ (x^2 -1) / sqrt(1-x^2) dx + ∫ (1/ sqrt(1-x^2)) dx

    So if you look the integral on the right is OK that takes care of the last integral in eq 2 in the attached pdf. But I'm stuck with this integral and I have tried literally everything I can think of to get it to the form -∫sqrt(1-x^2) dx from ∫ (x^2 -1) / sqrt(1-x^2) dx

    I tried to let u^2 = 1 - x^2 and sub back and sub the numerator for x^2 in terms of u.
    I tried to let u = 1 - x^2.
    I tried to let u = the numerator and I tried u^2 = numerator. I have tried trig substitution for the denominator I tried to draw a triangle and make sqrt ( 1 - x ^2) the adjacent side and proceed from there. NOTHING! This is frustrating because I have tried everything I know and I guarantee I will be mad because it is probably something simple I can't see. I feel I have exhausted my efforts

    Just look at the attached note for exercise 3. You will see.
    Thanks
     

    Attached Files:

  2. jcsd
  3. Jun 11, 2013 #2

    CAF123

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    Gold Member



    So what you have is: $$\int \frac{x^2}{\sqrt{1-x^2}}\,dx = \int \frac{x^2 - 1}{\sqrt{1-x^2}}\,dx + \int \frac{1}{\sqrt{1-x^2}}\,dx$$

    All they are doing to get the new form ##-\int \sqrt{1-x^2}\,dx## for the first term on the RHS is taking out a common factor on the numerator. What is that common factor?
     
  4. Jun 11, 2013 #3
    [itex]\int \frac{x^2}{\sqrt{1-x^2}}\,dx = \int \frac{x^2 - 1}{\sqrt{1-x^2}}\,dx + \int \frac{1}{\sqrt{1-x^2}}\,dx [/itex]


    Just a factor from this?
    [itex] I = \int \frac{x^2 - 1}{\sqrt{1-x^2}}\,dx [/itex]

    [itex] I = \int \frac{(x-1)(x+1)}{\sqrt{1-x^2}}\,dx [/itex]

    This is sad I don't see it.
     
  5. Jun 11, 2013 #4

    CAF123

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    Gold Member

    Okay, I'll start you off: (You'll be kicking yourself)
    $$x^2 - 1 = -(1-x^2)$$

    Now simplify.
     
  6. Jun 11, 2013 #5
    My foot hurts. Beyond dumb.
     
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