1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trig substitution integral

  1. Feb 4, 2013 #1
    1. The problem statement, all variables and given/known data
    [tex]\int \frac{dx}{\sqrt{x^{2}+16}}[/tex]


    2. Relevant equations



    3. The attempt at a solution
    [tex]x=4tan\theta[/tex] [tex]dx=4sec^{2}\theta d\theta[/tex]
    Therefore:
    [tex]\int \frac{4sec^{2}\theta d\theta}{\sqrt{16tan^{2}\theta +16}} = \int \frac{sec^{2}\theta d\theta}{\sqrt{tan^{2}\theta+1}}[/tex]
    [tex]\int \frac{sec^{2}\theta d\theta}{sec\theta} = \int sec\theta d\theta[/tex]
    [tex]=ln|sec\theta+tan\theta| = ln|sec\frac{\sqrt{x^{2}+16}}{4}+tan\frac{x}{4}|[/tex]

    Wolfram is getting a hyperbolic sine function so idk what is wrong (we've never talked about hyperbolic functions in class, so I don't think that's the answer)
     
  2. jcsd
  3. Feb 4, 2013 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You are OK at ##\ln|\sec\theta+\tan\theta|##. But in your very last expression you should leave out the secant and tangent, since you have calculated them in terms of x already. And you need to add a constant of integration. Then your answer will be correct.

    Don't worry about the hyperbolic functions. Integrals like this can be worked either way.
     
  4. Feb 4, 2013 #3
    Oh right (feel stupid lol). Are the hyperbolic functions an easier way to solve these problems?

    Thanks for the help
     
  5. Feb 4, 2013 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    They can be easier, but in this case I don't think it makes much difference. Try ##x = 4\sinh t## on it and see what you think. Remember the basic hyperbolic identity is ##\cosh^2t - \sinh^2 t = 1##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Trig substitution integral
Loading...