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Homework Help: How to integrate (1/x*sqrt(4x^2 -1)dx using parts or trig?

  1. Feb 24, 2007 #1
    1. The problem statement, all variables and given/known data
    (1 / x * sqrt(4x^2 - 1))dx


    2. Relevant equations
    done by parts/trig?


    3. The attempt at a solution
    1 = c
    2x = b
    sqrt(4x^2 - 1) = a

    sin(theta) = 2x

    sin(theta) / 2 = x
    -(1/2)*cos(theta) d(theta) = dx
    cos(theta) = -2*sqrt(4x^2 - 1)

    I am not sure what to do about the x?

    Can I just replace it all now?
     
    Last edited: Feb 24, 2007
  2. jcsd
  3. Feb 24, 2007 #2

    D H

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    The substitution you chose is not a particularly good one because [itex]4x^2-1[/itex] transforms to [itex]\sin^2\theta-1=-\cos^2\theta[/itex] with the substitution [itex]\sin\theta=2x[/itex], and [itex]-\cos^2\theta[/itex] is non-positive. (How do you take the square root?) You have [itex]cos\theta=-2\sqrt{4x^2-1}[/itex], which is incorrect.

    It is usually better to use some trigonometric substitution other than [itex]\sin\theta[/itex] when addressing equations involving [itex]ax^2-1[/itex].
     
    Last edited: Feb 24, 2007
  4. Feb 24, 2007 #3
    I have a triangle
    /_|

    I called the hypotenuse 1, the bottom leg sqrt(4x^2-1) and the right leg 2x. a = theta.

    sin(a) = 2x, x = sin(a) / 2, dx = cos(a) / 2 da
    sqrt(4x^2 - 1) = cos(a)

    replacing dx, x, and sqrt(16x^2 - 1).

    I get (cos(a)/2)da / sin(a)/2 * cos(a).

    I'm left with 1 / sin(a), integrate csc(a) = -ln[ csc(a) + cot(a) ];

    csc = 1 / 2x
    cot = sqrt(4x^2 -1) / 2x.

    -ln [ (1/2x) + (sqrt(4x^2 - 1) / 2x) ]

    Is that right now?

    I can't verify because if i plug it into "the integrator" I get arctan o_O
     
  5. Feb 24, 2007 #4

    D H

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    You are still using [itex]\sqrt{4x^2-1}=\cos a[/itex] with the substitution [itex]\sin a = 2x[/itex]. That is not correct. You are using the wrong substitution.
     
  6. Feb 24, 2007 #5
    I don't get what I can do =(.

    I looked at the table of integrals.

    du / u*sqrt(a^2 - u^2)

    spits out a - 1 / a * ln| [ a + sqrt(a^2 -u^2) ] / u | ?
     
  7. Feb 24, 2007 #6

    D H

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    http://en.wikipedia.org/wiki/Trigonometric_substitution" [Broken]
     
    Last edited by a moderator: May 2, 2017
  8. Feb 24, 2007 #7

    Gib Z

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    You want 2x to be the hypotenuse! Thats the way to get your radical to be one of the sides! If 1 is the hypotenuse, the radical is sqrt(1-something), and thats not what you want!
     
  9. Feb 24, 2007 #8

    Gib Z

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    /_| Hypot- 2x Right leg, radical, bot leg, 1, theta at bottom left
    [tex]\cos \theta = \frac{1}{2x}[/tex]

    [tex] \sec \theta = 2x[/tex]
    [tex]x=\frac{\sec \theta}{2}[/tex]
    [tex]dx=\frac{\sec \theta \tan \theta}{2} d\theta[/tex]
    [tex]\tan \theta=\sqrt{4x^2 -1}[/tex]

    You will get this last step after your correct substitutions.
    [tex]\int \frac{dx}{x\sqrt{4x^2-1}}= \theta[/tex]
     
    Last edited: Feb 24, 2007
  10. Feb 24, 2007 #9
    Holy Moly! That Is A Lot Cleaner! But what does theta = ?

    (theta) = -arctan(sqrt(4x^2-1))
     
    Last edited: Feb 24, 2007
  11. Feb 24, 2007 #10

    Gib Z

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    Ahh close, but no negative on the arc tan, since tan theta=sqrt(4x^2-1), arctan both sides,

    arc tan(tan theta)=arctan(sqrt4x^2-1)
    Which simplifies to

    theta=arctan(sqrt(4x^2-1))
     
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