What are examples of the integration trick involving "Simultaneous Integrals," "Dual Integrals," or "Pairs of Integrals"? I'm trying to get more apt at computing integrals, and I'm looking for examples of the particular integral technique of "Simultaneous Integrals," "Dual Integrals," or "Pairs of Integrals"?
Here (in Method 2) is an example of the technique I'm looking to understand. The technique is used as one option for integrating $\sqrt{\tan(x)}$, wherein two integrals $I$ and $J$ are introduced and the values of $I+J$ and $I-J$ are computed/used to determine the desired value.
My questions:

*

*What is this technique called? I included three names in my title; are any of those correct?

*(the most important) What are some other examples of using this technique in integration? The more tractable / lower-level, the better: I'd like to be able to convey this to Calculus students.

*(of lesser importance) What is the theoretical underpinnings of this technique that I need to be aware of? I've read that this is really a vector space basis situation at its core; is this really a linear algebra technique?

Down below, I collect some information:
Is this question a duplicate? Technically, yes. This 8-year-old question asks the same thing, but the answers there didn't really scratch the itch that I have.
What I've found independently:

*

*https://www.openbookpublishers.com/htmlreader/978-1-78374-142-7/Chapters/P26.html

*https://sci-hub.do/10.1080/10511971003742411

*https://www.jstor.org/stable/2974819
 A: The most famous one is probably one of the methods for working out the gaussian integral:
$$
I=\int\limits_{\mathbb R}e^{-x^2}dx=\int\limits_{\mathbb R}e^{-y^2}dy\\
\Rightarrow I^2=\left(\int\limits_{\mathbb R}e^{-x^2}dx\right)\left(\int\limits_{\mathbb R}e^{-y^2}dy\right)=\iint\limits_{\mathbb R^2}e^{-(x^2+y^2)}dx\,dy
$$
although this feels like cheating a bit as really it is the same integral twice

I guess another one would be something like this:
$$B=\int e^x\sin(x)dx\\A=\int e^x\cos(x)dx\\A+Bj=\int e^{(1+j)x}dx$$
which cuts a corner compared to using integration by parts
A: I don't really think the method you have in mind goes by any standard name though I have seen the phrase "a system of relations involving integrals" used.
Two indefinite examples include:

*

*Let
$$I = \int \frac{\sin^3 x}{\sin^3 x - \cos^3 x} \, dx \quad \text{and} \quad J = \int \frac{\cos^3 x}{\sin^3 x - \cos^3 x} \, dx.$$
If $I - J$ and $I + J$ are first found, $I$ and $J$ can then be found.

*Let
$$I = \int \frac{1 + x^4}{1 - x^4} \frac{dx}{\sqrt{1 + x^4}} \quad \text{and} \quad J = \int \frac{x^2}{1 - x^4} \frac{dx}{\sqrt{1 + x^4}}.$$
If
$$A = \int \frac{1 + x^2}{1 - x^2} \frac{dx}{\sqrt{ 1 + x^4}} \quad \text{and} \quad B = \int \frac{1 - x^2}{1 + x^2} \frac{dx}{\sqrt{1 + x^4}},$$
we see that
$$I = \frac{1}{2}(A + B) \quad \text{and} \quad J = \frac{1}{4}(A - B).$$
The integrals $A$ and $B$ can be found using, for example, a substitution $u = \sqrt{1 + x^4}/x$, from which $I$ and $J$ can then be found.

A few definite examples, at a slightly higher level of difficulty, include:

*

*Let
$$I = \int_0^{\frac{\pi}{4}} \log (\sin x) \, dx \quad \text{and} \quad J = \int_0^{\frac{\pi}{4}} \log (\cos x) \, dx.$$
Finding $I - J$ and $I + J$ leads to values for $I$ and $J$.

*Let
$$I = \int_0^1 \frac{x \log (1 - x)}{1 + x^2} \, dx \quad \text{and} \quad J = \int_0^1 \frac{x \log (1 + x)}{1 + x^2} \, dx.$$
Again finding $I - J$ and $I + J$ leads to values for $I$ and $J$.

*Let
$$I = \int_0^1 \arctan (x) \log (1 - x) \, dx \quad \text{and} \quad J = \int_0^1 \arctan (x) \log (1 + x) \, dx.$$
Again finding $I - J$ and $I + J$ leads to values for $I$ and $J$.

A: A related example to this method seems the following. Let $f:[-1,1]\to \mathbb{R}$ be even and continuous, and let $c\in \mathbb{R}$.
Consider
$$
I:=\int_{-1}^1 \frac{f(x)}{1+\exp(cx)} dx.
$$
The substitution $y=-x$ yields
$$
I=\int_{-1}^1 \frac{f(y)}{1+\exp(-cy)} dy.
$$
Thus
$$
I+I=\int_{-1}^1f(x) \left(\frac{1}{1+\exp(cx)}+\frac{1}{1+\exp(-cx)}\right)  dx
$$
$$
=\int_{-1}^1f(x) dx = 2\int_{0}^1f(x) dx.
$$
So (independent on $c$)
$$
I=\int_{0}^1f(x) dx.
$$
For example
$$
\int_{-1}^1 \frac{x^4}{1+\exp(x)} dx= \frac{1}{5}.
$$
It is not exactly the method, as only the sum and not the difference of integrals is used, but it seems related.
