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Seemingly simple integral gets complicated

  1. Oct 21, 2005 #1
    Well, I'm an AP Calculus BC student, and I always liked to get ahead, so before the year started, I went through Calc I and II. Now that I'm in the class with a 100+, I'm simply refreshing myself on the Calc II portion. My brother decided to give me a seemingly simple problem that turned out more complicated than I expected. Any assistance would be greatly appreciated.
    Problem: Evaluate integral
    [tex]\int \sin{x} \sqrt{(\cos{x})^2 + 1}dx[/tex]
    I decided to use substitution initially, setting u = cos x. dx = du/-sin(x), thus eliminating the sin(x) and leaving [tex]\int \sqrt{u^2 + 1}du [/tex]
    I then thought my best bet would be to approach this with trignometric substitution. I said tan([tex]\theta[/tex]) = u so
    [tex]\sec{\theta}^2d\theta = du [/tex]
    [tex]\int (\sec{\theta})^2\sqrt{(\tan(\theta)} +1)dx [/tex]
    I then evaluated [tex]\sqrt{ (\tan{\theta})^2 +1 } = (\sec{\theta})[/tex]
    Thus, my integral was simplified (relatively) down to
    [tex] \int (\sec{\theta})^3 d\theta [/tex]
    It just gets worse from there. Am I making some horrid mistake in my trignometric substitution or before or have I completely approached this incorrectly? I have more work that I've done, but I want to ensure that this is correct so far. Any help would be great. Thanks
     
    Last edited: Oct 21, 2005
  2. jcsd
  3. Oct 21, 2005 #2

    Fermat

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    try u = sinht
     
  4. Oct 21, 2005 #3
    It looks a-ok so far. Just don't forget the minus sign in your first u-substitution.

    Do you know how to integrate secant cubed?
     
  5. Oct 22, 2005 #4

    HallsofIvy

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    Your brother is a smart-ass!

    To integrate [itex]sec^3(\theta)d(\theta)[/itex], first write it as [itex]\frac{d\theta}{cos^3(\theta)}[/itex]. Now multiply numerator and denominator by [itex]cos(\theta)[/itex] to get [itex]\frac{cos(\theta)d\theta}{cos^4(\theta)}= \frac{cos(\theta)d\theta}{(1-sin^2(\theta)^2}[/itex].
    Let [itex]y= sin(\theta)[/itex] and this becomes [itex]\frac{dy}{(1-y^2)^2}[/itex] which can be done by partial fractions.
     
  6. Oct 22, 2005 #5
    When I was in calculus I was taught that the only way to integrate secant cubed was by parts. Obviously this is not the case :rolleyes: .

    By parts does not take too long, if you remember what to choose for u and dV!
     
  7. Oct 22, 2005 #6
    To elaborate on what Fermat said, the integral can be written as:

    [tex]\int\sqrt{1+x^2}\,dx[/tex]

    Then you can use the suggested substitution.

    And speaking of parts, try using dv = sec2θ.
     
    Last edited: Oct 22, 2005
  8. Oct 22, 2005 #7
    Thanks to everyone who replied. For some reason, I didn't think about attempting integration by parts.

    Thanks to suggestsions everyone, I'll try them out. A question though, for the y = sinh(t), if I'm correct, is that hyperbolic trignometric substitution? I'm not very familiar with hyperbolic functions. I'll read up on it. Thanks to everyone. Once I get some more work done on it, I'll post again to see if I did it correctly.
     
  9. Oct 22, 2005 #8

    Fermat

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    Yes, that's a hyperbolic substitution. I goes like this,

    [tex]\int\ \sqrt{u^2 + 1}\ du[/tex]

    [tex]\mbox{let } u = sinh\ t\mbox{, then,}[/tex]

    [tex]du = cosh\ t\ dt[/tex]

    carrying out the substitution,

    [tex]\int\ \sqrt{sinh^2\ t + 1}\cdot cosh\ t\ dt[/tex]
    [tex]\int\ \sqrt{cosh^2\ t}\cdot cosh\ t\ dt[/tex]
    [tex]\int\ cosh^2\ t\ dt[/tex]

    [tex]\mbox{There is an ordinary trig identity, } cos^2 x = \frac{1}{2} (1 + cos2x)[/tex]
    all you have to do now is find the equivalent hyperbolic trig identity (try google), substitute, integrate, and back-substitute for x in the original integral.
     
    Last edited: Oct 22, 2005
  10. Oct 22, 2005 #9
    Fascinating: I've never seen hyperbolic trignometry used for substitution. Thank you for elucidating on that. I''ll go study up on that.
     
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