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Trig Sub or what?

  1. Apr 22, 2007 #1
    1. The problem statement, all variables and given/known data
    integral of (16x-2x^2-23)^1/2 dx


    2. Relevant equations



    3. The attempt at a solution
    First off it seems clear that I must complete the square within the radical.
    This gives ((x-4)^2-4.5)^1/2

    From there I seem to be lost, fumbling through my book for any kind of help. Can I use a trig integration, such as using the form (a^2-x^2)^1/2 to substitute x=asecT? I don't think so because (x-4)^2 cannot be treated like x^2.

    I also cannot find any possible u-sub that will work because of the radical over the entire problem.

    Any type of guidance is very much appreciated!
     
  2. jcsd
  3. Apr 22, 2007 #2

    siddharth

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    Gold Member

    Set x-4 as t, and integrate wrt t.
     
  4. Apr 22, 2007 #3
    Some work...

    integral signs before everything

    a=sqrt(4.5)
    x-4 = asecT
    dx=asecTtanTdT


    sqrt[-2*(asecT)^2-4.5] * sqrt[4.5] * secTtanTdT

    sqrt[-2*(4.5sec^2(T)-4.5)*4.5] * secTtanTdT

    sqrt[(-40.5)(tan^2(T)] * secTtanTdT

    Subbing T=arcsec((x-4)/a)

    sqrt[(-40.5)(tan^2(arcsec(T))] * sec(arcsec(T))tan(arcsec(T))dT

    ..heres where I start to get lost
    I know sec(arcsec(t)) is just T.
    I know you can do a case for tan(arcsec(T))..
    if x>=1 =) sqrt(x^2-1)
    or x<1 =) -sqrt(x^2-1)

    though I don't think that helps much.

    I'm very lost and desperate for some help, any advice is appreciated. Thank you.
     
  5. Apr 22, 2007 #4

    Dick

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    Science Advisor
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    Try putting (x-4)=a*sin(T). Since sin^2(T)-1=-cos^2(T) this will take care of the awkward negative sign under the integral and make your life overall much easier.
     
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