- #1
HotMintea
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1. The problem statement
Use dimensional analysis to find \int\sqrt{\ a\ - \ b\ x^2\ }\ dx.
A useful result is \int\sqrt{\ 1\ - \ x^2\ }\ dx\ = \frac{arcsin{x}}{2}\ + \frac{x\sqrt{\ 1\ - \ x^2\ }}{2}\ + \ C.
2. The attempt at a solution
If I let <b> = L^2 </b> and [x] = M, then [a] = L^2 M^2 and [\int\sqrt{\ a\ - \ b\ x^2\ }\ dx]\ = LM^2.
Hence, my answer was:
<br /> \begin{equation*} <br /> \begin{split} <br /> \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.<br /> \end{split} <br /> \end{equation*}<br />
However, the correct answer (by Wolfram Alpha) was:
<br /> \begin{equation*} <br /> \begin{split} <br /> \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.<br /> \end{split} <br /> \end{equation*}<br />
( 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.
Use dimensional analysis to find \int\sqrt{\ a\ - \ b\ x^2\ }\ dx.
A useful result is \int\sqrt{\ 1\ - \ x^2\ }\ dx\ = \frac{arcsin{x}}{2}\ + \frac{x\sqrt{\ 1\ - \ x^2\ }}{2}\ + \ C.
2. The attempt at a solution
If I let <b> = L^2 </b> and [x] = M, then [a] = L^2 M^2 and [\int\sqrt{\ a\ - \ b\ x^2\ }\ dx]\ = LM^2.
Hence, my answer was:
<br /> \begin{equation*} <br /> \begin{split} <br /> \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.<br /> \end{split} <br /> \end{equation*}<br />
However, the correct answer (by Wolfram Alpha) was:
<br /> \begin{equation*} <br /> \begin{split} <br /> \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.<br /> \end{split} <br /> \end{equation*}<br />
( 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.
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