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

Homework Help: Integral Problem (very Hard)

  1. Mar 4, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove that [tex]\int[/tex][tex]\sqrt{}9-x^2[/tex] dx

    =[tex]\frac{9\Theta}{2}[/tex]+[tex]\frac{9sin2\Theta}{4}[/tex]+c

    given that x=3sin[tex]\Theta[/tex]
    2. Relevant equations



    3. The attempt at a solution
    [tex]\int[/tex][tex]\sqrt{}9-x^2[/tex] dx

    =[tex]\frac{(9-x^2)^{1.5}}{10x}[/tex]

    =[tex]\frac{(-x^2)(\sqrt{-x^2+9}+9\sqrt{(-x^2)+9}}{10x}[/tex]

    =[tex]\frac{-x(x^2\sqrt{(-x^2)+9}-9(\sqrt{(-x^2)+9}}{10}[/tex]

    =[tex]\frac{-3sin\Theta(9(sin^2)\Theta\sqrt{9(sin^2)\Theta-9}-9\sqrt{9(sin^2)+9}}{10}[/tex]


    Im not to sure if im going in the right direction if i am not guidance would be appreciated
     
  2. jcsd
  3. Mar 4, 2010 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You're not going in the right direction. Your very first step is wrong. The power rule only applies when the integrand is of the form xndx where [itex]n \ne -1[/itex]. It doesn't apply when you have some function of x taken to a power, as you do in this case. Also, I have no idea where that 10x in the denominator came from.

    Use the substitution given and rewrite the integral in terms of [itex]\theta[/itex] first.
     
  4. Mar 4, 2010 #3
    ok so if i do...

    [tex]\int[/tex][tex]\sqrt{9-9sin^2\Theta}[/tex]
    =[tex]\int[/tex]3-3sin[tex]\Theta[/tex]
    =3[tex]\Theta[/tex]+3cos[tex]\Theta[/tex]

    is it going in the right direction now?? thanks for the help
     
    Last edited: Mar 4, 2010
  5. Mar 4, 2010 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Marginally better. First of all,

    [tex]\sqrt{a^2-b^2} \ne a-b[/tex]

    Second, you forgot the dx and then didn't write it in terms of [itex]d\theta[/itex]. Finally, not that it really matters, you didn't integrate the first term correctly.

    I would suggest you review your textbook on the topic of trig substitutions. There's probably a similar example you could use as a template for solving this problem.
     
  6. Mar 4, 2010 #5

    HallsofIvy

    User Avatar
    Science Advisor

    [itex]\sqrt{9- 9 sin^2(\theta)}= 3\sqrt{1- sin^2(\theta)}= 3\sqrt{cos^2(\theta)}[/itex]
     
  7. Mar 4, 2010 #6
    okay i was finally able to prove it!!

    [tex]\int\sqrt{a^2-x^2}[/tex] dx
    =[tex]\int a^2-a^2sin^2\Theta[/tex] acos[tex]\Theta[/tex] d[tex]\Theta[/tex]
    =[tex]\int\sqrt{a^2(1-sin^2\Theta}[/tex] acos[tex]\Theta[/tex] d[tex]\Theta[/tex]
    =[tex]\int\sqrt{a^2cos^2\Theta}[/tex] acos[tex]\Theta[/tex] d[tex]\Theta[/tex]
    =[tex]\int a^2cos^2\Theta[/tex] d[tex]\Theta[/tex]

    =[tex]\frac{a^2}{2}[/tex]([tex]\Theta[/tex]+sin[tex]\Theta[/tex]cos[tex]\Theta[/tex])

    i know i definatly skipped a couple of steps in integrating cos^2 :redface:
     
    Last edited: Mar 4, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook