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Calculus BC Trigonometric Substitition

  1. Nov 11, 2009 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    [tex]\sin^2{\theta}+\cos^2{\theta}=1[/tex]

    3. The attempt at a solution

    [tex]x=\tan{\theta}[/tex]

    I simplified it down to:

    [tex]\int \frac {1}{\sin{\theta} \cdot \cos^2{\theta}} \, d\theta[/tex]

    Which I do not know how to integrate.

    Any help would be wonderful!
     
    Last edited: Nov 11, 2009
  2. jcsd
  3. Nov 11, 2009 #2
    Never mind... I rewrote it as:

    [tex]\int \csc{\theta}\cdot\sec^2{\theta} \, d\theta[/tex]

    And used u-substition:

    [tex]u=\sec{\theta}[/tex]

    [tex]\int \frac{\csc{\theta} \cdot \sec^2{\theta}}{\tan{\theta} \cdot \sec{\theta}} \, du [/tex]

    [tex]\int \csc^2{\theta} \, du[/tex]

    [tex]\int \frac{u^2}{u^2-1} \, du[/tex]

    [tex]\frac{1}{2} \cdot \ln |\frac{u-1}{u+1}| +u+C [/tex]

    I then simplified using my previous substitutions and logarithm rules to get my final answer:

    [tex] \ln |\frac{\sqrt{1+x^2}-1}{x}|+\sqrt{1+x^2}+C [/tex]
     
  4. Nov 11, 2009 #3

    Mark44

    Staff: Mentor

    And you can check by differentiating your answer. If you end up with sqrt(1 + x^2)/x, you're golden.
     
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