Emmanuel José García's user avatar
Emmanuel José García's user avatar
Emmanuel José García's user avatar
Emmanuel José García
  • Member for 1 year, 8 months
  • Last seen more than 1 year ago
About

The half-angle formulas are central!

Context: The theoretical importance of half-angle formulas

And as a picture is worth a thousand words...

enter image description here

A transformation that simplifies certain composite trigonometric integrals and offers an alternative approach to integrating through trig./hyp. substitution.

$$\int f\left(x,\tan{\frac{\beta}{2}}, \tan{\frac{\gamma}{2}} \right)\,dx=\int f\left(\frac{e^{i\alpha}+e^{-i\alpha}}{2}a, e^{i\alpha}, \frac{1-e^{i\alpha}}{1+e^{i\alpha}}\right)\,\frac{e^{-i\alpha}-e^{i\alpha}}{2i}a\,d\alpha\tag{1}$$

Where $\alpha=\cos^{-1}\left(\frac{x}{a}\right)$, $\beta=\csc^{-1}\left(\frac{x}{a}\right)$, $\gamma=\sec^{-1}\left(\frac{x}{a}\right)$ and $a>0$.

$$\int f\left(x, \sqrt{x^2-a^2}, \frac{\sqrt{x-a}}{\sqrt{x+a}}\right)\,dx= \int f\left(\frac{e^{i\alpha}+e^{-i\alpha}}{2}a, \frac{e^{-i\alpha}-e^{i\alpha}}{2}a, \frac{1-e^{i\alpha}}{1+e^{i\alpha}}\right)\,\frac{e^{-i\alpha}-e^{i\alpha}}{2i}a\,d\alpha\tag{2}$$

$$\int f\left(x, \sqrt{x^2+a^2}\right)\, dx = \int f\left(\frac{e^{\theta}-e^{-\theta}}{2}a, \frac{e^{\theta}+e^{-\theta}}{2}a\right) \frac{e^{\theta}+e^{-\theta}}{2}a\, d\theta\tag{3}$$

Where $\theta=\sinh^{-1}(\frac{x}{a})$ and $a>0$

Discovery is the privilege of the child: the child who has no fear of being once again wrong, of looking like an idiot, of not being serious, of not doing things like everyone else. - A. Grothendieck

Badges
This user doesn’t have any gold badges yet.
This user doesn’t have any silver badges yet.
1
bronze badge
Posts

This user hasn’t posted yet.