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

How to [itex]\int{\frac{-2x+20}{2\sqrt{-x^2+20x}}}dx[/itex]

  1. Jun 5, 2015 #1
    Hello,

    What would be a good method, for a novice in integration, to solve this integral. I haven't done much integration, so this looks quite daunting. Would this be solvable with trig substitution by itself or would I need to employ partial fractions?

    [itex]\int{\frac{-2x+20}{2\sqrt{-x^2+20x}}}dx[/itex]

    If I complete the square in the denominator, I wonder if I could make it easier.

    [itex]\int{\frac{-2x+20}{2\sqrt{-(x-10)^2-10}}}dx[/itex]

    What would be a logical method?
     
  2. jcsd
  3. Jun 5, 2015 #2

    phion

    User Avatar
    Gold Member

    Try a u-substitution.
     
  4. Jun 5, 2015 #3

    phion

    User Avatar
    Gold Member

    [itex]\frac{1}{\sqrt{u}}[/itex]
     
  5. Jun 6, 2015 #4
    Hi,

    I found a similar example in the Stewart calculus book, chapter 7.3 example 7. I completed the square in the denominator, did u-substitution u=x-10, then trig substitution u= 10sin(theta).

    The integratable term I arrived at was [itex] \int (-10sin(theta)) = 10cos(theta)[/itex]

    Then I substituted back in the x terms to get [itex]y(x)=\sqrt{10^2 -(x-10)^2)} => ((x-10)^2)+y^2=10^2[/itex]

    Which is a circle with radius 10 and I believe with center at the x axis and left side passing through the origin.

    I think this is correct as it is the equation I initially started out with and took the derivative of before integrating back.

    (note: I apologize, I don't have the latex avalable on my phone. I did my best and will fix any errors when I get to a computer at school in a few hours.)
     
  6. Jun 6, 2015 #5

    phion

    User Avatar
    Gold Member

    You over-complicated things a little bit, but nice job!
     
  7. Jun 6, 2015 #6
    You can also use u-substitution as phion said


    ##
    \int\frac{-2x+20}{\sqrt{-x^2+20}}dx
    ##

    As you can see

    -2x+20 is derivative of -x^2+20

    And if you substitute -x^2+20 with variable such as t, then the problem becomes simpler

    ##
    t=-x^2+20x
    ##

    ##
    \frac{d(t)}{dt} = \frac{d(-x^2+20x)}{dt}
    ##

    ##
    1 = (-2x+20) \cdot \frac{dx}{dt}
    ##

    ##
    dt = (-2x+20) \cdot dx
    ##

    You can substitue ##(-2x+20) \cdot dx## with ##dt##
    and ##-x^2+20## with ##t##

    then

    ##
    \int\frac{-2x+20}{\sqrt{-x^2+20}}dx
    ##

    becomes

    ##
    \int\frac{1}{2\sqrt{t}} dt
    ##

    ##
    =\frac{1}{2}\int t^{-\frac{1}{2}} dt
    ##

    ##
    =t^{\frac{1}{2}}
    ##
     
  8. Jun 7, 2015 #7
    Thank you.

    Wow, that was a much more efficient choice of substitution. There must be many choices for substituting in the integrand. Most of which are overly complicated and a few of which are more elegant. Thanks for shedding light on this topic.

    Regards
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How to [itex]\int{\frac{-2x+20}{2\sqrt{-x^2+20x}}}dx[/itex]
  1. Int of cos(u^2) (Replies: 23)

Loading...