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Integration using Trig Substitution

  1. Feb 8, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\int \frac{cosx dx}{\sqrt{1 + sin^{2}x}}[/tex]

    2. Relevant equations

    Expression: [tex]\sqrt{a^{2} + x^{2}}[/tex]
    Substitution: x = a*tan[tex]\Theta[/tex]
    Identity: 1 + tan[tex]^{2}\Theta = sec^{2}\Theta[/tex]

    3. The attempt at a solution
    I have tried using Trig Substitution, but I end up getting an equation much like the one I started only it contains secants and tangents instead of cosine and sine. For some reason I only see a circular route that would just take me back to the original equation.
     
    Last edited: Feb 8, 2008
  2. jcsd
  3. Feb 8, 2008 #2

    cristo

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    Try the substitution y=sin(x). That should then put the integrand in a familiar form.
     
  4. Feb 8, 2008 #3
    as in changing the bottom from 1 + (sinx)^2 to 1 + y^2 ?


    I can get to:

    [tex]\int \frac{sec^{2}\Theta d\Theta}{\sqrt{1+tan^{2}\Theta}}[/tex]

    But after that it makes no sense, I just get back to the original equation when I sub. again.
     
    Last edited: Feb 8, 2008
  5. Feb 8, 2008 #4

    Gib Z

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    After y=sin x, and then a tan theta = y, you should have the same integrand with a plus sign, not a minus. Then use the Pythagorean identities to simplify the square root.
     
  6. Feb 8, 2008 #5
    So I should end up with:

    [tex]\int sec\Theta d\Theta}[/tex]

    Correct?

    then I would get ln [tex]\left| sec\Theta+tan\Theta \right|[/tex] + C
     
    Last edited: Feb 8, 2008
  7. Feb 8, 2008 #6

    cristo

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    If you use the substitution I suggested, then you end up with [tex]\int\frac{dy}{\sqrt{1+y^2}}[/tex]. This is a standard integral, is it not?
     
  8. Feb 8, 2008 #7
    Yes, I'm fairly certain that is a standard integral, and it comes out to:

    [tex]sin^{-1}y + C[/tex]

    But I'm also fairly certain that I am supposed to solve this problem using the expression, substitution, and identity when I do the trig sub. Maybe I am overcomplicated the problem and should just stick to the arcsin answer, Gib Z's method seems to be the one I'm after
     
    Last edited: Feb 8, 2008
  9. Feb 8, 2008 #8

    cristo

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    Close; the solution is [tex]\sinh^{-1} y +C[/tex]
     
  10. Feb 8, 2008 #9
    My mistake, you're right. Thanks!
     
    Last edited: Feb 8, 2008
  11. Feb 8, 2008 #10

    Gib Z

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    It's always nice to know alternative methods, because we can know express the inverse hyperbolic sine function in terms of the natural logarithm =] Which we could have also done more directly, but o well!
     
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