miércoles, 17 de julio de 2024
Showing $\frac{1}{e^{i\alpha_2}} + \frac{1}{e^{i\beta_2}} + \frac{1}{e^{i\gamma_2}} = \frac{1}{e^{i(\alpha_2+\beta_2+\gamma_2)}}$
viernes, 5 de julio de 2024
Integrals of the form $\int f\left(x,\frac{\sqrt{x+m}}{\sqrt{x+n}}\right)\,dx$
Let $b-a=m$ and $b+a=n$, where $a$ and $b$ are real numbers. Then the following identities hold:
$$\frac{1-e^{\pm\text{i}\alpha}}{1+e^{\pm\text{i}\alpha}}=\tan{\left(\frac12\sec^{-1}\left(\frac{x+b}{a}\right)\right)}=\frac{\sqrt{x+b-a}}{\sqrt{x+b+a}}=\frac{\sqrt{x+m}}{\sqrt{x+n}},\tag{1}$$
$$e^{\pm\text{i}\alpha}=\tan\left(\frac{1}{2} \csc^{-1}\left(\frac{x+b}{a}\right)\right) = \frac{x + b - \sqrt{(x + b)^2 - a^2}}{a}\tag{2}$$
These identities are generalizations of identities $(1-2)$ and $(9-10)$ at "Integration using Euler-like identities" and can be derived similarly as shown in the linked blog. The general identities $(1)$ and $(2)$ lead us to the following general transformation formula:
$$\boxed{\int f\left(x,\tan{\frac{\beta}{2}}, \tan{\frac{\gamma}{2}} \right)\,dx=\int f\left(\frac{e^{i\alpha}+e^{-i\alpha}}{2}a, e^{\pm\text{i}\alpha}, \frac{1-e^{\pm\text{i}\alpha}}{1+e^{\pm\text{i}\alpha}}\right)\,\frac{e^{-\text{i}\alpha}-e^{\text{i}\alpha}}{2i}a\,d\alpha}\tag{3}$$
Where $\alpha=\cos^{-1}\left(\frac{x+b}{a}\right)$, $\beta=\csc^{-1}\left(\frac{x+b}{a}\right)$ and $\gamma=\sec^{-1}\left(\frac{x+b}{a}\right).$ Use the upper sign for the alternating signs $\mp$ when $\frac{x + b}{a} \geq 1$ and the lower sign when $0 \leq \frac{x + b}{a} \leq 1$.
To illustrate its usefulness, let's look at the following example taken from the YouTube channel Prime Newtons.
Example 1. Evaluate
$$\int \frac{\sqrt{x+1}}{\sqrt{x+2}}\,dx$$
Solution. First, we obtain the values of $a$ and $b$ by solving the following (simple!) system of equations:
$$\left. b-a=1\atop a+b=2\right\}$$
The solutions are $a = \frac{1}{2}$ and $b = \frac{3}{2}$. Applying formula $(3)$ for $x\geq-1$, the integral becomes
$$\begin{aligned}\int \frac{\sqrt{x+1}}{\sqrt{x+2}}\,dx&= -\frac{i}{4}\int \frac{(1 - e^{i\alpha}) (e^{-i\alpha} - e^{i\alpha})}{(e^{i \alpha} + 1)} \, d\alpha\\&=-\frac{i}{4}\int \frac{e^{-i\alpha} (e^{i\alpha} - 1) (e^{2i\alpha} - 1)}{e^{i\alpha} + 1} \, dx\\&=-\frac{i}{4}\int e^{-i\alpha}(e^{i\alpha}-1)^2\,d\alpha\\&=-\frac{i}{4}\int (e^{-i\alpha}+e^{i\alpha}-2)\,d\alpha\\&=\frac14(2i\alpha+e^{-i\alpha}-e^{i\alpha})+C\end{aligned}$$
To switch to real numbers, replace from $(2)$, then replace the values of $a$ and $b$ and simplify, obtaining
$$=\sqrt{x^2+3x+2} + \frac{1}{2} \log\left(2x - 2\sqrt{x^2+3x+2} + 3\right)+C$$
On the Prime Newtons YouTube channel, they first perform a variable change, followed by a trigonometric substitution, reducing the integral to integrals involving the secant and secant cubed. This integral calculator uses $u^2 = \frac{x+1}{x+2}$ and then (a complicated) partial fraction decomposition. This can be even more complicated for integral like this one:
$$\int \frac{x\sqrt{x+1}}{\sqrt{x+2}}\,dx$$
where as our method only requires a few lines of algebra and basic calculus, as in Example 1.