# Dimensional analysis for arcsine integral

• HotMintea
In summary, the conversation discussed the problem of using dimensional analysis to find the integral of a given equation and the differences in the attempted solution and the correct answer provided by Wolfram Alpha. The conversation also touched on the concept of arctan and its relationship to the given equation.
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 $[\int\sqrt{\ a\ - \ b\ x^2\ }\ dx]\ = LM^2$.

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

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:
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).

ideasrule said:
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).

That is true! Thanks for your help!

## 1. What is dimensional analysis for arcsine integral?

Dimensional analysis for arcsine integral is a mathematical method used to analyze and solve integrals involving arcsine functions. It involves manipulating the units of measurement in the integral and using them to determine the appropriate substitution to make the integral dimensionless.

## 2. Why is dimensional analysis important for solving arcsine integrals?

Dimensional analysis is important because it allows us to simplify complex integrals and reduce them to dimensionless form, making them easier to solve. It also ensures that the final solution has the correct units of measurement.

## 3. How do you perform dimensional analysis for arcsine integrals?

To perform dimensional analysis for arcsine integrals, you need to identify the variables and their units of measurement in the integral. Then, you need to choose an appropriate substitution that will eliminate the units and make the integral dimensionless. This can involve using trigonometric identities or other mathematical techniques.

## 4. What are some common mistakes to avoid when using dimensional analysis for arcsine integrals?

One common mistake is using an incorrect substitution that does not eliminate the units or makes the integral more complex. It is also important to ensure that the final solution has the correct units of measurement, as this indicates a successful dimensional analysis.

## 5. Can dimensional analysis be used for other types of integrals?

Yes, dimensional analysis can be used for other types of integrals, such as logarithmic, exponential, and trigonometric integrals. It is a useful tool for simplifying and solving a wide range of mathematical problems.

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