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Integral: sqrt(4-x^2)*sign(x-1) dx

  1. Dec 26, 2009 #1
    Integral: sqrt(4-x^2)*sign(x-1) dx

    Can someone help me with this integral?
    I've never worked with a sign funtion before, so I have absolutely no idea how to solve this integral
     
  2. jcsd
  3. Dec 26, 2009 #2

    HallsofIvy

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    Science Advisor

    I would be inclined to do this as two separate integrals.

    If [itex]x\ge 1[/itex], [itex]x-1\ge 0[/itex] so sng(x-1)= 1 and your integral is [itex]\int \sqrt{4-x^2}dx[/itex].

    If x< 1, x-1< 0 so sgn(x-1)= -1 and your integral is [itex]-\int\sqrt{4- x^2}dx[/itex].
     
    Last edited by a moderator: Dec 28, 2009
  4. Dec 26, 2009 #3
    Thanks, but how do you solve [itex]\int \sqrt{4-x^2}dx[/itex]
    I've tried with: x= 2sin(t) ==> dx= 2cos(t)dt and sin(t)= x/2 ==> t= arcsin(x/2)

    But then I get: 2 arcsin(x/2) + sin(2(arcsin(x/2)))
    And that's wrong: the first part is correct, but the second part isn't.
    What have I done wrong?
     
  5. Dec 26, 2009 #4

    Mentallic

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    [tex]\int \sqrt{4-x^2}dx[/tex]

    Now, using your substitutions...

    [itex]x=2sint[/itex]
    [itex]dx=2costdt[/itex]

    [tex]\int \sqrt{4-(2sint)^2}.(2costdt)[/tex]

    Simplifying this would give you:

    [tex]4\int cos^2tdt[/tex]

    Did you get this? Now, you'll need to use a double-angle equality for the trigo to take the integral of this.
     
  6. Dec 26, 2009 #5
    I solved it till there..
    but what I did then was wrong

    This is what I did:

    [tex]4\int cos^2tdt[/tex]

    = 4((t/2)+(sin(2t)/4))
    = 2t + sin (2t)

    Because t= arcsin(x/2) you get:

    2(arcsin(x/2)) + sin(2arcsin(x/2))

    The first part is correct, the second isn't..
    What have I done wrong?
     
  7. Dec 28, 2009 #6
    I solved it till there..
    but what I did then was wrong

    This is what I did:

    [tex]4\int cos^2tdt[/tex]

    = 4((t/2)+(sin(2t)/4))
    = 2t + sin (2t)

    Because t= arcsin(x/2) you get:

    2(arcsin(x/2)) + sin(2arcsin(x/2))

    The first part is correct, the second isn't..
    What have I done wrong?
     
  8. Dec 28, 2009 #7

    uart

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    Not wrong, just not fully simplified. Often when you have a trig of inverse-trig you can simplify it to an algebraic expression. In this case start out be using :

    [tex] \sin(2 \theta) = 2 \sin(\theta) \, \cos(\theta)[/tex]

    and see if you can simplify it from there.

    BTW. Since you know it's "wrong" then I assume that you have the correct answer at hand and you know that this second part should simplify to :

    [tex] \frac{x}{2} \, \sqrt{4-x^2} [/tex]

    It does. :)
     
  9. Dec 28, 2009 #8
    Thank you very much!
    But.. when I do this (t= arcsin(x/2)) I get:

    sin(2t)
    = 2 sin(t) cos (t)
    = 2 sin(arcsin(x/2)) cos(arcsin(x/2)
    = 2 (x/2) (sqrt(1-(x/2)))

    Do I have to bring the 2 into the sqrt to get sqrt(4-x^2)?
     
  10. Dec 28, 2009 #9

    HallsofIvy

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    You are missing the square in [itex]2(x/2)\sqrt{1- (x/2)^2}[/itex].
    You don't have to but putting that 2 inside the square root will give you the form of the answer above:
    [tex](x/2)\sqrt{4(1- x^2/4)}= (x/2)\sqrt{4- x^2}[/tex]
     
  11. Dec 28, 2009 #10
    Thank you!
     
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