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Dimensional analysis for arcsine integral

  1. Jul 4, 2011 #1
    1. The problem statement

    Use dimensional analysis to find [itex] \int\sqrt{\ a\ - \ b\ x^2\ }\ dx [/itex].

    A useful result is [itex] \int\sqrt{\ 1\ - \ x^2\ }\ dx\ = \frac{arcsin{x}}{2}\ + \frac{x\sqrt{\ 1\ - \ x^2\ }}{2}\ + \ C [/itex].

    2. The attempt at a solution

    If I let [itex] = L^2 [/itex] and [itex] [x] = M [/itex], then [itex] [a] = L^2 M^2 [/itex] and [itex] [\int\sqrt{\ a\ - \ b\ x^2\ }\ dx]\ = LM^2 [/itex].

    Hence, my answer was:

    [tex]
    \begin{equation*}
    \begin{split}
    \int\sqrt{\ a\ - \ b\ x^2\ }\ dx\ = \frac{a}{\sqrt{b}}\frac{arcsin{\frac{\sqrt{b}\ x}{\sqrt{a}}}}{2}\ + \frac{x\sqrt{\ a\ - \ b\ x^2\ }}{2}\ + \ C.
    \end{split}
    \end{equation*}
    [/tex]

    However, the correct answer (by Wolfram Alpha) was:

    [tex]
    \begin{equation*}
    \begin{split}
    \int\sqrt{\ a\ - \ b\ x^2\ }\ dx\ = \frac{a}{\sqrt{b}}\frac{arctan{\frac{\sqrt{b}\ x}{\sqrt{a\ - \ bx^2\ }}}}{2}\ + \frac{x\sqrt{\ a\ - \ b\ x^2\ }}{2}\ + \ C.
    \end{split}
    \end{equation*}
    [/tex]
    ( http://www.wolframalpha.com/input/?i=int+sqrt%28a-bx^2%29dx [/URL])

    I wonder why [itex] \int\sqrt{\ a\ - \ b\ x^2\ }\ dx\ [/itex] will not be the same as [itex] \int\sqrt{\ 1\ - \ x^2\ }\ dx\ [/itex] when a = b = 1. Moreover, I would like to know how to find [itex] arctan{\frac{\sqrt{b}\ x}{\sqrt{a\ - \ bx^2\ }}} [/itex] part by dimensional analysis or similar method without doing the full integral.
     
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Jul 4, 2011 #2

    ideasrule

    User Avatar
    Homework Helper

    They're two different ways of writing the same thing. If opposite/hypotenuse for a certain angle is √b*x over √a, opposite/adjacent would be √b*x over √(a-bx^2).
     
  4. Jul 4, 2011 #3
    That is true! Thanks for your help! :smile:
     
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