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: Trigonometric integral

  1. Mar 18, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\int\frac{1}{t^3\sqrt{t^2-1}}dt[/tex] with limits of integration [tex][\sqrt{2}, 2][/tex]

    2. Relevant equations



    3. The attempt at a solution

    Using trig. sub, I have [tex]sec \theta=t[/tex]

    [tex]dt=sec \theta tan \theta d \theta[/tex]

    [tex]\int\frac{1}{t^3\sqrt{t^2-1}}dt[/tex] with limits of integration [tex][\sqrt{2}, 2][/tex]

    [tex]=\int\frac{sec \theta tan\theta d \theta}{(sec \theta)^3\sqrt{(sec \theta)^2-1}}[/tex]

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

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

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

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

    [tex]=\int\cos \theta^2 d\theta[/tex]

    [tex]=\int\frac{1+cos \theta}{2} d\theta[/tex]

    [tex]=\int\frac{1}{2}+\frac{cos \theta}{2} d\theta[/tex]

    [tex]=\frac{1}{2}\int d \theta +\frac{1}{2}\int cos \theta d \theta[/tex]

    [tex]=\frac{1}{2} \theta | +\frac{1}{2}sin \theta |[/tex]

    [tex]=\frac{1}{2}arcsec t +\frac{1}{2}t[/tex] with limits [tex][\sqrt{2},
    2][/tex]

    [tex]=\frac{1}{2}[arcsec (2)-arcsec (\sqrt{2})]+\frac{1}{2}[2-\sqrt{2}][/tex]

    I need to get to the answer: [tex]\frac{\pi}{24}+\frac{\sqrt{3}}{8}-\frac{1}{4}[/tex]. I don't see anything wrong up to this point so I guess my question is more of an algebra question, but how could I arrive at the stated answer?
     
  2. jcsd
  3. Mar 18, 2010 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    [tex]\cos^2\theta = \frac{1 + \cos{(2 \theta})}{2}[/tex]
     
  4. Mar 19, 2010 #3
    That was a silly mistake.

    Alright, so:

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

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

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

    [tex]=\frac{1}{2}\theta| +\frac{1}{2}\frac{sin(2\theta)}{2}|[/tex]

    [tex]=\frac{1}{2}\theta| +\frac{1}{4}sin(2\theta)|[/tex]

    [tex]=\frac{1}{2}\theta| +\frac{1}{4}(2sin\theta cos\theta)|[/tex]

    [tex]=\frac{1}{2}arcsin (t)| +\frac{1}{2}(\frac{t^2-1}{t})(\frac{1}{t})|[/tex]

    [tex]=\frac{1}{2}arcsin (t)| +\frac{1}{2}(\frac{t^2-1}{t^2})|[/tex]

    [tex]=\frac{1}{2}arcsin (t)| +\frac{1}{2}(1-\frac{1}{t^2})|[/tex]

    with limits of integration [tex][\sqrt{2}, 2][/tex]

    [tex]=\frac{1}{2}(arcsin(2)-arcsin(\sqrt{2})) +\frac{1}{2}[(1-\frac{1}{2^2})-(1-\frac{1}{\sqrt{2}^2})[/tex]

    [tex]=\frac{1}{2}(arcsin(2)-arcsin(\sqrt{2})) +\frac{1}{2}[(1-\frac{1}{4})-(1-\frac{1}{2})][/tex]

    [tex]=\frac{1}{2}(arcsin(2)-arcsin(\sqrt{2})) +\frac{1}{2}[(1-\frac{1}{4})-(1-\frac{1}{2}})][/tex]

    [tex]=\frac{1}{2}(arcsin(2)-arcsin(\sqrt{2})) +\frac{3}{8}-\frac{1}{4}[/tex]

    I'm not sure if this is correct, but how do I deal with the arcsin values since they aren't in the domain of the standard arcsin function? Lastly, how would I algebraically convert this to
    [tex]
    \frac{\pi}{24}+\frac{\sqrt{3}}{8}-\frac{1}{4}
    [/tex]?

    Thanks.
     
  5. Mar 19, 2010 #4

    ideasrule

    User Avatar
    Homework Helper

    You don't have to go through all this drama. You know that t=sec(theta), so cos(theta)=1/t. Just use this to convert the limits of integration to theta.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook