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Integral of x(sqrt(1-x^2)) + 1/2 -x^2/2 IS IT RIGHT?

  1. May 5, 2009 #1
    1. The problem statement, all variables and given/known data

    integral of [ x*sqrt[1-x2] + 1/2
    - x2/2] dx - with the limits of 1 and 0

    the last two bits are easy but i just wanted to confirm what i have done for the first bit is correct...ive tried using a method that takes less time but i dont know if its right...

    2. Relevant equations

    sin(x)^2 + cos(x)^2 = 1

    cos(arcsinx) = sqrt(1-x2)

    3. The attempt at a solution

    for the first bit:

    x = sinu
    dx = cosu du
    1-x2 = cos(u)^2

    so it becomes: sinu(cos(u)2)cosu du

    sinu du = -dcosu

    so :- -cos(u)3 dcosu
    integrate this beacomes: -cos(u)4/4

    which then becomes
    -cos(arcsinx)4/4 which is just: -(1-x2)2/4

    so putting this together with the 2nd and 3rd part gives

    -(1-x2)2/4 + x/2 - x3/6 with limts 1&0

    which equals 1/3

    so is this right?
     
  2. jcsd
  3. May 5, 2009 #2

    Cyosis

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    Use the right tool for the right situation. Trigonometric substitution certainly works, but it takes you on quite a detour. To check if your answer is correct you simply differentiate it so lets do that.

    [tex]\frac{d}{dx} \frac{-(1-x^2)^2}{4}=x(1-x^2) \neq x \sqrt{1-x^2}[/tex]

    You make an error very early on after you state that 1-x^2=cos(u)^2 you conclude in the next line that [itex]\sqrt{1-x^2}=\cos^2u[/itex]. In other words you forgot to take the square root. The rest seems to be correct.

    There is however a method that is way easier take u=1-x^2 and you're done.
     
  4. May 5, 2009 #3
    as soon as you pointed out that error early on i felt like kicking myself :rolleyes: yeah i see what to do now

    (do you still get 1/3 as the answer cause ive just redone it and im still getting the same answer with the limits in)

    thanks
     
    Last edited: May 5, 2009
  5. May 5, 2009 #4

    Cyosis

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    For the first term I get 1/3 for the second 1/2 and for the third -1/6. So 1/3+1/2-1/6=2/3
     
  6. May 5, 2009 #5

    tiny-tim

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    Welcome to PF!

    Hi indie452! Welcome to PF! :smile:

    (have a square-root: √ :wink:)
    indie452, have you tried Cyosis's :smile: sub, u = √(1 - x2)?

    It's much quicker, and it's a standard technique that you do need to learn. :smile:
     
  7. May 5, 2009 #6
    yeah i tried it ... the reason i got the wrong answer the 2nd time was cause i messed up on subbing the limits but i got the right answer now
     
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