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Integration using Trig. Substitution

  1. Feb 1, 2010 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution

    I used the substitution X=tan [tex]\theta[/tex]

    So, [tex]dX=(sec^2 \theta[/tex]) d[tex]\theta[/tex]

    Substituting in for X, I get:

    [tex]\int\sqrt{(tan^2 \theta)+1}(sec^2 \theta) d\theta[/tex]

    = [tex]\int\sqrt{(sec^2 \theta)}(sec^2 \theta) d\theta[/tex]

    = [tex]\int(sec \theta)(sec^2 \theta) d\theta[/tex]

    I then converted secants into cosines:

    = [tex]\int\frac{1}{(cos \theta)(cos^2 \theta)} d\theta[/tex]

    = [tex]\int\frac{1}{(cos \theta)(1-sin^2 \theta)} d\theta[/tex]

    I then used U-sub:

    [tex]u=sin \theta[/tex]

    [tex]du=cos \theta[/tex]

    [tex]\frac{du}{cos \theta}=d\theta[/tex]

    = [tex]\int\frac{du}{(cos^2 \theta)(1-u^2)}[/tex]

    = [tex]\int\frac{du}{(1-u^2)(1-u^2)}[/tex]

    = [tex]\int\frac{du}{(1-2u^2+u^4)}[/tex]

    = [tex]\int\frac{du}{(u^4-2u^2+1)}[/tex]

    = [tex]\int\frac{du}{u^2(u^2-2)+1}[/tex]

    I see this:

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

    I was hoping I could then use substitution and have the integral of [tex]arctan[/tex], but it looks much more complicated than I thought. Am I anywhere near the right track?
  2. jcsd
  3. Feb 1, 2010 #2
    At that point, you could attempt partial fractions. However, an easier way is here:

    \int(sec \theta)(sec^2 \theta) d\theta = \int sec^3 \theta d\theta

    This is a famous (sorta, anyways) integral. Use parts twice.

    Alternatively, since this integral is so famous, it even has its own wikipedia page.


    I'd suggest you trying the derivation yourself before looking at the "answer" but it's there if you need it.
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