# Magdy's Exchanger

## Theorem

Magdy's Exchanger is a simple method to get a semi-algebraic formula for inverse trigonometric functions using the help of inverse hyperbolic functions that has a real logarthmic formula hence we will have an example to get the inverse sine function.

First:

$\displaystyle \arcsin x = \int \frac 1 {\sqrt{1 - x^2}} \, \mathrm d x$

Using Integration by Parts:

$\displaystyle \int \frac 1 {\sqrt{1-x^2}} \,\mathrm d x = -\frac {\sqrt {1-x^2}} x - \int \frac{\sqrt{1-x^2}} {x^2} \, \mathrm d x$

The problem is that the integration on the right hand side will inverse itself if we solve it by parts.

Then the result will be:

$0 = 0$

Magdy's method will come into effect now.

As we know the inverse hyperbolic function that is most similar to the inverse sine function is the inverse sinh function.

The idea is to get the last integration in the right hand side from the inverse sinh function.

So we use parts again on the derivative of inverse sinh as follows:

$\displaystyle \operatorname{arcsinh} x = \int \frac 1 {\sqrt{1 + x^2}} \, \mathrm d x$

 $\displaystyle \operatorname{arcsinh} x$ $=$ $\displaystyle \int \frac 1 {\sqrt{1 + x^2} } \, \mathrm d x$ Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$ $\displaystyle$ $=$ $\displaystyle \frac {\sqrt {1 + x^2} } x + \int \frac {\sqrt {1 + x^2} } {x^2} \,\mathrm d x$ Integration by Parts $\displaystyle$ $=$ $\displaystyle \ln \left({x + \sqrt{x^2 + 1} }\right)$ Definition of Inverse Sinh Function
$\displaystyle \int \frac {\sqrt {1 + x^2} } {x^2} \, \mathrm d x = \ln \left({x + \sqrt{x^2 + 1} }\right) - \frac {\sqrt {1 + x^2} } x$

Taking the derivativeof both sides:

$\displaystyle \frac{\sqrt{1 + x^2} } {x^2} = \frac {1 + x^2 + x \sqrt{x^2 + 1} } {x^3 + x^2 \sqrt {x^2 + 1} } = A$

Squaring both sides:

$\dfrac {1 + x^2} {x^4} = A^2$

Now the most important step.

As we want to have $1 - x^2$ in the numerator we will have the fraction on the left hand side} as:

$\dfrac {1 - x^2 + 2 x^2} {x^4}$

which does not affect the value of the fraction.

Now the left hand side is:

$\dfrac {1 - x^2} {x^4} + \dfrac 2 {x^2}$

then the wanted fraction will be alone on the left hand side:

$\dfrac {1 - x^2} {x^4} = A^2 - \dfrac 2 {x^2}$

Taking the root of both sides:

$\dfrac {\sqrt{1 - x^2} } {x^2} = \sqrt {A^2 - \dfrac 2 {x^2} }$

then integrating both sides:

$\displaystyle \int \frac {\sqrt{1 - x^2} } {x^2} \, \mathrm d x = \int \sqrt{A^2 - \frac 2 {x^2}} \, \mathrm d x$

Now the right hand side is complicated but evaluates as follows:

$\displaystyle \int \sqrt{A^2 - \frac 2 {x^2} } \, \mathrm d x = \frac {x^2 \sqrt {\frac 1 {x^4} - \frac 1 {x^2} } \ln \left({2 \sqrt {x^2 - 1} + 2 x}\right)} {\sqrt {x^2 - 1} } - x \sqrt {\frac 1 {x^4} - \frac 1 {x^2} } + C$

Substituting the last result in the first equation of the inverse sine we get:

$\displaystyle \arcsin x = - \frac {\sqrt {1 - x^2} } x - \frac {x^2 \sqrt {\frac 1 {x^4} - \frac 1 {x^2} } \ln \left({2 \sqrt {x^2 - 1} + 2 x}\right)} {\sqrt {x^2 - 1}} + x \sqrt {\frac 1 {x^4} - \frac 1 {x^2} } + C$

Hence we get a semi-algebraic formula for the inverse sine function.

$\blacksquare$

### Notes

1) From the result it's obvious that we call the outputs "semi-algebraic" formula because it has a logarithm, but it's a real logarithm, so the is the purpose of that method is to get the inverse trigonometric functions in terms of a real formulas or functions like the real logarithm in our result.

2) In the step that we get $\dfrac{\sqrt{1 - x^2}} {x^2}$ from the form $\dfrac{\sqrt{1 + x^2}}{x^2}$ it's recommended to add $-x^2 + x^2$ to the numerator under the root so that the plenty in the left side after squaring $\dfrac 1 {x^2}$ will be added to the right side "of course with an inverse sign" that will give an easier quantity in the right side than if we multiply the left side by $\dfrac{\sqrt{1 - x^2}}{\sqrt {1 - x^2}}$ that will result in a more complex quantity in the right side and that doesn't have integration.

3) Furthermore,we observe from solution that Magdy's Exchanger is a method to get an easier form for the derivative of inverse trigonometric functions that can be integrated using computers.

4) by the same way and putting the last notes into consideration we can get a semi-algebraic formulas for the other inverse trigonometric functions.

## Source of Name

This entry was named for Ahmed Magdy.

Ahmed Magdy is an Egyptian engineering student.