User:Ahmed Magdy Hosny/sandbox

Magdy's Exchanger "named after an egyption engineering student Ahmed Magdy "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 we have the fact that :

$$\ arcsin x = \int\frac{1}{\sqrt{1-x^2}}\, dx $$

using integration by parts method we get :


 * $$ \int\frac{1}{\sqrt{1-x^2}}\,dx = -\frac{\sqrt{1-x^2}}{x}\ - \int\frac{\sqrt{1-x^2}}{x^2}\,dx $$

now the problem is the integration in the right side that will inverse itself if we solve it by parts then the result will be 0 = 0 so Magdy's method will affect now. As we know the most inverse hyperbolic function that is similar to the inverse sine function is the inverse sihn function and the idea is to get the last integration in the right side from inverse sihn function so we will use parts again on the derivative of inverse sihn as the following :


 * $$\operatorname{arsinh}\,x =\int\frac{1}{\sqrt{1+x^2}}\,dx $$


 * $$\int\frac{1}{\sqrt{1+x^2}}\,dx =\frac{\sqrt{1+x^2}}{x}+\int\frac{\sqrt{1+x^2}}{x^2}\,dx $$

and from the inverse sihn logarthmic form we have :


 * $$\operatorname{arsinh}\,x = \ln(x + \sqrt{x^2+1})$$

so the integration of the right side will be :

$$\int\frac{\sqrt{1+x^2}}{x^2} dx =  \ln(x + \sqrt{x^2+1}) - \frac{\sqrt{1+x^2}}{x} $$

taking the derivative of the both sides will lead to :

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

squaring the both sides we get :

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

now the most important step. As we want to have  $$  1-x^2  $$    in the numerator we will have the fraction on the left side as :

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

which dosen's affect the value of the fraction. Now we will have the left side as :

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

then we will have the wanted fraction alone in the left side so :

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

now taking the root of both sides :

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

then taking the integration of the both sides :

$$\int\frac{\sqrt{1-x^2}}{x^2} dx = \int\sqrt{A^2-\frac{2}{x^2}} dx $$

now the right side is complex and we can use computer to calculate it that will give :

$$\int\sqrt{A^2-\frac{2}{x^2}}dx = \frac{x^2\sqrt{\frac{1}{x^4}-\frac{1}{x^2}}\ln(2\sqrt{x^2-1}+2x)}{\sqrt{x^2-1}}- x\sqrt{\frac{1}{x^4}-\frac{1}{x^2}}+C $$

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

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

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

===important 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   $$\frac{\sqrt{1-x^2}}{x^2}$$    from the form    $$\frac{\sqrt{1+x^2}}{x^2}    $$ 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    $$\frac{1}{x^2}$$     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    $$\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}}$$   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-algabraic formulas for the other inverse trigonometric functions.