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Integration with squareroot

  1. Sep 15, 2007 #1
    I can't seem to figure out how to integrate this.

    integ ( sqrt (1+x^2) dx )

    I am pretty sure I need to substitute the inside but I cannot figure out what. I have tried substituting x = tan theta, and i get stuck at integral (sec (theta)^3 dtheta)

    here is my work: (I am goin to use q as theta)

    integ ( sqrt (1+x^2) dx )
    let tan q = x
    sec^2q dq = dx

    integ ( sqrt (1+tan^2q) sec^2 q dq )
    = integ (sqrt (sec^2 q) sec^2 q dq ) (trig identity)
    = integ (sec q (sex^2 q) dq)
    = integ (sec^3 q dq)
    = integ (1/sin^3 q dq)

    and I am stuck here.....
    unless this is the wrong substitution, in which case, i have no clue.
  2. jcsd
  3. Sep 15, 2007 #2
    Try [tex]\int{\sec{x}\sec^2{x}dx} = \int{\sec{x}\left(1+\tan^2{x}\right)dx}[/tex]

    And [itex]\sec{x}[/itex] is [itex]\frac{1}{\cos{x}}[/itex], not [itex]\frac{1}{\sin{x}}[/itex]. :wink:
  4. Sep 15, 2007 #3


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    Substitute [itex]x=\sinh t [/itex]. And use a double angle formula for the hyperbolic functions.
  5. Sep 15, 2007 #4
    You can also do it with integration by parts, it's pretty lengthy but you don't have to deal with hyperbolic or trig substitutions.
  6. Sep 15, 2007 #5

    Gib Z

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    Dex's substitution gets you there the fastest, but you may not know about hyperbolic trig functions yet. You can use the x= tan theta substitution, but of course the resulting sec cubed integral is not the easiest. So if you really want to go that way, do integration by parts on the sec cubed integral, its actually not that lengthy compared to other elementary methods. You should get the sec cubed integral on both sides of the equation, then its just simple algebra.
  7. Sep 17, 2007 #6
    I have tried to do the hyperbolic trig way because I am stuck on mine and I can't seem to integrate the sec^3q

    this is what i have for hyperbolic, i am stuck and can't seem to integrate the last one

    integ (sqrt (1+x^2)dx)
    let x = sinh t
    dx/dt = cosh t
    integ (sqrt (1+sinh^2t) (cosh t dt))
    identity: cosh^2t - sinh^2t = 1
    integ (sqrt (cosh^2t) cosh t dt)
    integ (cosh^2t dt)

    how do i integrate this?
  8. Sep 17, 2007 #7

    Gib Z

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    [tex]\cosh^2 x = \frac{1 + \cosh (2x)}{2}[/tex]
  9. Sep 17, 2007 #8
    okay, so I have done everything and integrated the whole thing, now because I have never learnt about the hyperbolic function, i am having trouble doing this.

    here is my work for the rest of the integration:
    (previous work)
    integ (sqrt (1+x^2)dx)
    let x = sinh t
    dx/dt = cosh t
    integ (sqrt (1+sinh^2t) (cosh t dt))
    identity: cosh^2t - sinh^2t = 1
    integ (sqrt (cosh^2t) cosh t dt)
    integ (cosh^2t dt)
    (finishing it off)
    integ ( (1+ cosh(2t))/2 dt)
    = 1/2 integ ( 1+cosh(2t) dt)
    = 1/2 [ t + 1/2 sinh (2t) ]

    So my actual question was to evaluate this:
    and now with the sinh, I am unsure how to evaluate this. What is the relationship between sinh and sin?

    actual question:
  10. Sep 19, 2007 #9

    Gib Z

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    Well to do your iterated integral:

    First finish off the anti derivative for the inner integral. To do this, first use the definition:
    [tex]\sinh x = \frac{e^x - e^{-x}}{2}[/tex]. We define the inverse function and call it argsinh. Also, use the following identitys (which you can verify from the definition): [tex]\sinh (2t) = 2 \sinh t \cosh t[/tex] and [tex]\cosh^2 t - \sinh^2 t = 1[/tex]

    So now the anti derivative is [tex]\frac{ \argsinh x}{2} + \frac{x \sqrt{1+x^2}}{2}[/tex]

    Now, argsinh 1 = 0, because if you let x=0 into the sinh definition, you get 1.
    The Inner integral is now thus: [tex]\frac{\sqrt{2}}{2} - \frac{ \argsinh y}{2} - \frac{y\sqrt{1+y^2}}{2}[/tex].

    So now for the outer integral, the first part is very straightforward, the 3rd part just needs a substitution u=y^2+1. The only one that is somewhat difficult is the argsinh one, which you can get by integration by parts. First take the factor of 1/2 out of the integral, then let u= argsinh y, dv=dy.

    Good luck
  11. Sep 19, 2007 #10
    okay, i have read this about 10 times trying to understand this, i have tried substituting everything in to get the antiderivative you got, but i constantly get something different.
    why is this integral so hard , am i doing it the hard way is there an easier way to compute this.
  12. Sep 19, 2007 #11


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    Using some techniques, it's pretty easy.

    Attached Files:

  13. Sep 19, 2007 #12
    i'm not sure if i can use the formula to help the integration because i also have this formula somewhere, but i don't see how else i can integrate without it as the solution with the hyperbolic substitution is too difficult for me.

    nevermind i see how you did this, and i would have never known how to do this...
    i clearly am in the wrong major
    Last edited: Sep 19, 2007
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