- #1

HotMintea

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**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.

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.

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