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Another tri integration

  1. Apr 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Integral of 6/(sqrt(4-x^2))

    2. Relevant equations

    arcsin(u) = u'/ sqrt(1-u^2)

    3. The attempt at a solution

    WA gave a different answer
    http://www.wolframalpha.com/input/?i=Integral+of+6%2F%28sqrt%284-x^2%29%29

    I got the following:

    Integral of 6/(sqrt(4-x^2))
    divide by 4 to reduce to 1-u^2, and pull out the constant 4: 1/4 int of 6/sqrt(1-(x/2)^2
    i was going to pull out the six right away, but i could integrate that with du.
    u = x/2, du = 1/2 dx, so 2 du = dx
    i get 12 inside, and pull out the 12, i get 12/4 int u' / sqrt(1-u^2)
    which gives me 3 arcsin(x/2)+c

    but this is not right from WA.

    Thank you.
     
  2. jcsd
  3. Apr 7, 2010 #2
    U sub
    x=2sin(u),
    dx=2cos(u)du.
    It'll become
    6*2cos(u)/((4-4(sin(u))^2)^(1/2)
     
  4. Apr 7, 2010 #3

    Mark44

    Staff: Mentor

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

    I got the same as WA. This is a very simple trig substitution. I always draw a right triangle, since I don't like to clutter up my head with a bunch of formulas if I don't have to. The hypotenuse is 2, the opposite side is x, and the adjacent side is sqrt(4 - x^2).

    From this I get 2sinw = x, and 2cosw dw = dx, and 2 cos w = sqrt(4 - x^2)

    [tex]\int \frac{6 dx}{\sqrt{4 - x^2}} = 6 \int \frac{2cos w dw}{2cos w} = 6\int dw = 6w + C[/tex]

    From 2sin w = x, I get sin w = x/2 ==> w = sin-1(x/2), so the antiderivative is 6sin-1(x/2) + C.

    YOU KEEP FORGETTING YOUR DX! IT WILL COME AROUND TO BITE YOU ONE DAY!
     
  5. Apr 7, 2010 #4
    Hi. Thank you for doing these for me.
    I think I have learned the mistake.
    All the sub are corected except when I pulled out the 1/4 , i didn't square root it. It should be 1/2 outside.

    LOL Thanks. I just left dx out since there was only one variable, namely, x.

    But I WILL REMEMBER THAT...
     
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