Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Integration using Trig. Substitution

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

    [tex]\int\sqrt{X^2+1}dX[/tex]

    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:

    [tex]
    \int(sec \theta)(sec^2 \theta) d\theta = \int sec^3 \theta d\theta
    [/tex]

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

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

    http://en.wikipedia.org/wiki/Integral_of_secant_cubed

    I'd suggest you trying the derivation yourself before looking at the "answer" but it's there if you need it.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook