"Complexification formulas are great and it seems like this simplifies the right away."Ninad Munshi
If complex $\theta_1=\cos^{-1}(p)$ and $\theta_2=\sec^{-1}(p)$, where $p\in(-1, 0) \cup (1, \infty)$, then the following relation holds:
$$e^{i\theta_1}=\frac{1-\tan\frac{\theta_2}{2} }{1+\tan\frac{\theta_2}{2}}.
\tag{1}\label{463459_1}$$
And for $p\in(-\infty, -1)\cup (0, 1)$, we have
$$e^{-i\theta_1}=\frac{1-\tan\frac{\theta_2}{2} }{1+\tan\frac{\theta_2}{2}}. \tag{2}\label{463459_2}$$
If complex $\theta_1=\sin^{-1}(p)$ and $\theta_2=\csc^{-1}(p)$, where $p\in(-\infty, -1)\cup (0, 1)$, then the following relation holds:
$$ie^{i\theta_1}=-\tan\frac{\theta_2}{2}.\tag{3}\label{463459_3}$$
And for $p\in(-1, 0) \cup (1, \infty)$ we have
$$ie^{-i\theta_1}=\tan\frac{\theta_2}{2}.\tag{4}\label{463459_4}$$
There are several variants that we can obtain by equating (and simplifying) the trigonometric formulas for quadratic equations from my previous question.
I have noticed that for certain trigonometric integrals defined over permissible intervals of $p$, the evaluation simplifies considerably. For instance, consider the following definite integral:
$$\int_2^5 \sqrt{\tan\left(\frac{\csc^{-1}(x)}{2}\right)} \,dx.\tag{5}$$
This integral calculator returns the following (although Wolfram Alpha solves it):
No antiderivative could be found within the given time limit, or all supported integration methods were tried unsuccessfully. Note that many functions don't have an elementary antiderivative.
But it gives an approximation of $1.178881841955109.$
Given that the interval $[2, 5]$ is within the permissible values of $p$, we can use $e^{i\cos^{-1}(p)}=\tan\frac{\csc^{-1}(p)}{2}$ (derived from identities $(1)$ and $(4)$, valid for $p\in[-1, 0) \cup [1, \infty)$, to convert $(5)$ into
$$\int_2^5 \sqrt{e^{i\cos^{-1}(x)}}\,dx.\tag{6}$$
The same calculator provides the same answer but now displaying the steps as well. As a second example, Mathematica is unable to solve this integral:
$$\int_{2}^{3} \frac{{1 - \tan\frac{{\sec^{-1}x}}{2}}}{{1 + \tan\frac{{\sec^{-1}x}}{2}}}\sqrt{\tan\frac{\csc^{-1}x}{2}}\,dx\tag{7}$$
Neither this integral calculator. Although both the calculator and Wolfram Alpha can give you a numerical approximation. However, thanks to this new trick, you can convert $(7)$ into
$$\int_{2}^{3} e^{\frac{3i\arccos(x)}{2}}\,dx\tag{8}$$
Note that this integral calculator has no problem solving (elegantly!) integral $(8)$.
Other examples of integrals that at least Wolfram Alpha is not capable of solving but that can be evaluated using the transformations described in this blog (each integral is linked to its solution in the integral calculator):
So far I have considered integrals involving $\tan{\left(\frac12\csc^{-1}x\right)}$. However, this technique can be applied to a countless number of cases that surpass my initial expectations. For example, consider the integral $\int_{\frac{\pi}{4}}^{\frac{\pi}{20}}\sqrt{\tan{x}}\,dx$. This can be solved by letting $x=\frac12\csc^{-1}t$, where $\,dx=-\frac{1}{2t\sqrt{t^2-1}}\,dt$, transforming the original integral into $-\frac12\int_{\csc(\frac{\pi}{2})}^{\csc(\frac{\pi}{10})}e^{\frac12i\arccos(t)} \left(\frac{1}{t\sqrt{t^2-1}}\right) \, dt$. Certainly, there will be instances where employing these transformations might seem overly intricate, akin to cutting bread with a saw. However, what I aim to emphasize is the remarkable versatility of this technique.
Related material
No hay comentarios:
Publicar un comentario