# Homework Help: Dimensional analysis for arcsine integral

1. Jul 4, 2011

### HotMintea

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 $= L^2$ and $[x] = M$, then $[a] = L^2 M^2$ and $[\int\sqrt{\ a\ - \ b\ x^2\ }\ dx]\ = LM^2$.

Hence, my answer was:

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

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

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

I wonder why $\int\sqrt{\ a\ - \ b\ x^2\ }\ dx\$ will not be the same as $\int\sqrt{\ 1\ - \ x^2\ }\ dx\$ when a = b = 1. Moreover, I would like to know how to find $arctan{\frac{\sqrt{b}\ x}{\sqrt{a\ - \ bx^2\ }}}$ part by dimensional analysis or similar method without doing the full integral.

Last edited by a moderator: Apr 26, 2017
2. Jul 4, 2011

### ideasrule

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).

3. Jul 4, 2011

### HotMintea

That is true! Thanks for your help!

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook