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!

Integrate w/negative exponent

  1. Apr 15, 2007 #1
    I need a little 'suggestion' as to how to integrate cos^6x sin^-3x dx.

    I rewrite to cos^6x/sin^3x dx and let u = sinx but when I'm trying to rewrite integral, what should I do with the ^6?

    Thanks!
     
  2. jcsd
  3. Apr 15, 2007 #2
    oh wait, re-reading the problem and I think what I'm interpreting as a negative is the dot in the i in sin.... I'm going to work the problem as cos^6x sin^3x dx and I know how to work this... sorry about that!
     
  4. Apr 16, 2007 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    For future reference, if you did have an odd power of sine in the denominator, multiply both numerator and denominator by sine:
    [tex]\int \frac{cos^6 t}{sin^3 t}dt= \frac{cos ^6 t sin t}{sin^4 t}dt[/tex]
    [tex]= \int \frac{cos^6 t}{(1- cos^2 t)^2}(sin t dt)[/tex]
    so that the substitution x= cos t gives
    [tex]-\int\frac{x^6}{(1-x^2)^2}dx[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Integrate w/negative exponent
Loading...