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!

Using the substitution u=1/x

  1. Jul 17, 2007 #1

    danago

    User Avatar
    Gold Member

    Using the substitution u=1/x, evaluate:

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

    I was able to do it making the substitution [tex]x=cos\theta[/tex], but im supposed to show a worked solution using the given substitution.

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

    Thats about as far as i was able to get.

    Any help? :redface:

    Thanks,
    Dan.
     
  2. jcsd
  3. Jul 17, 2007 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Don't stop there. Replace x with 1/u.
     
  4. Jul 17, 2007 #3
    Hint: if [tex]u=1/x[/tex], what is [tex]du[/tex] ?

    OPPS: I see you already got that.

    Now, complete the substitution process in you last integral ( [tex]x = 1/u[/tex] ) and simplify. You'll then find that another simple substitution will yield further progress.
     
    Last edited: Jul 17, 2007
  5. Jul 17, 2007 #4

    danago

    User Avatar
    Gold Member

    Ahh i think i see what to do. Doing what Dick said, i ended up with:

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

    From that, it looks like i can make the sub [tex]u=sec \theta[/tex], which ill go and try now :smile:
     
  6. Jul 17, 2007 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Easier yet. Try v=u^2-1.
     
  7. Jul 17, 2007 #6

    danago

    User Avatar
    Gold Member

    oh yea ofcourse. We just went through trig subs in class, so thats all ive been thinking of trying now lol :yuck:

    Thanks very much for the help :approve:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Using the substitution u=1/x
Loading...