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The following identities have been suggested based on formulas in a previous question of mine.

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:

$$\int_2^5 \sqrt{\tan\left(\frac{\csc^{-1}(x)}{2}\right)} \,dx.\tag{5}\label{463459_5}$$

This integral calculator returns the following:

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 $p$, I can use $e^{i\cos^{-1}(p)}=\tan\frac{\csc^{-1}(p)}{2}$ (derived from identities \eqref{463459_1} and \eqref{463459_4}), valid for $p\in[-1, 0) \cup [1, \infty)$, to convert \eqref{463459_5} into

$$\int_2^5 \sqrt{e^{i\cos^{-1}(x)}}\,dx.\tag{6}\label{463459_6}$$

The same calculator provides the same answer but now displaying the steps as well. As a second example, compare the solutions of the following definite integrals, coming from identity \eqref{463459_1}:

$$\int_2^3 \sqrt{\dfrac{2}{\tan\left(\frac{\operatorname{arcsec}(x)}{2}\right)+1}-1} \, dx\tag{7}\label{463459_7}$$

$$\int_2^3 e^{\frac{i\arccos(x)}{2}} \,dx\tag{8}\label{463459_8}$$

Judge for yourself which one is simpler to solve between \eqref{463459_7} and \eqref{463459_8}.

I have spent quite some time applying these substitutions and have come to the conclusion that these identities make certain trigonometric integrals less tedious. This leads me to the question: Was this trick known in the world of integrals?

EDITED. Not exactly the same technique , but I think this is somehow related to my question. According to Wikipedia, using Euler's formula, any trigonometric function may be written in terms of complex exponential functions, namely $e^{ix}$ and $e^{-ix}$ and then integrated. This technique is often simpler and faster than using trigonometric identities or integration by parts, and is sufficiently powerful to integrate any rational expression involving trigonometric functions.

Crossposted at MO.

This story will continue on my blog.

Emmanuel José García
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    Complexification formulas are great and it seems like this simplifies the right away. You just have to be careful that when you draw triangles and take square roots the signs make sense at every step, especially the end, to line up with your original integrand. – Ninad Munshi Feb 03 '24 at 16:24
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    The simplication in real space is convenient as well $$\tan\frac{\csc^{-1}x}2=x-\sqrt{x^2-1}$$ – Quanto Feb 03 '24 at 17:20
  • @Quanto: Yes, although I must say that the substitution I have used is not the best. The substitution $$e^{i\cos^{-1}(x)}=\tan{\frac{\csc^{-1}(x)}{2}},$$ valid for $x\in[-1, 0) \cup [1, \infty)$ is simpler. – Emmanuel José García Feb 03 '24 at 20:49
  • Rewriting in terms of radicals becomes tedious in other, more complicated cases. I prefer to stick with the complexification formulas, as Ninad has called them. – Emmanuel José García Feb 03 '24 at 22:23
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    Another example of a complexification formula is treating $$\int\cos(\log x)dx = \operatorname{Re}\left{\int x^i dx\right}$$ in case you haven't come across the term before. It comes from a variety of places, one such prominent elementary example being complexifying vector spaces to add pairwise dimensional structure to go from a $2n$ to $n$ dimensional space. – Ninad Munshi Feb 04 '24 at 06:58
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    Why do we have to integrate $\sqrt{e^{i\arccos(x)}}$ in order to evaluate $\int_2^5 \sqrt{\tan\left(\frac{\csc^{-1}(x)}{2}\right)} ,dx$? Why can't we just use (4) to write $i e^{-i\arcsin(x)}=\tan\left(\frac{\csc^{-1}(x)}{2}\right)$ and then integrate $\sqrt{i e^{-i\arcsin(x)}}$? – David K Feb 05 '24 at 19:00
  • @DavidK: Because I realized that this transformation is simpler. The closed form is also simpler than using $(4)$. – Emmanuel José García Feb 05 '24 at 20:00
  • Wolfram Alpha says equation $(1)$ is not true for $0 < p < 1$, but instead equation $(2)$ is true. It says neither is true at $p=-1$ (the value on the right-hand side is undefined). Remind me how you decided that equation $(1)$ is true for all $p \geq -1$? – David K Feb 23 '24 at 01:55
  • @DavidK: I decided empirically. I'm writing a note and still pondering some details. Tomorrow I'll review the constraints of $p$ and edit my question. Thank you. – Emmanuel José García Feb 23 '24 at 03:16
  • With Mathematica: $$\int_2^5 \sqrt{\tan \left(\frac{1}{2} \csc ^{-1}(x)\right)} , dx=\int_2^5 \sqrt{\frac{1}{x+\sqrt{-1+x^2}}} , dx=-\frac{1}{3} \left(7 \sqrt{2}\right)+4 \sqrt{3}-\sqrt{6}$$ – Mariusz Iwaniuk Feb 23 '24 at 16:40
  • @MariuszIwaniuk: Yes, but check the comments here. Although I still need to correct the restrictions on $p$ in MO. – Emmanuel José García Feb 23 '24 at 16:51
  • @DavidK: Consider the identity $e^{i\cos^{-1}(x)}=\tan\frac{\csc^{-1}(x)}{2}$. The exponential $e^{i\cos^{-1}(x)}$ can be rewritten as $x+i\sqrt{1-x^2}$ and $\tan\frac{\csc^{-1}(x)}{2}$ as $x-\sqrt{x^2-1}$ (as Quanto mentioned above). Assume $x$ is real. Then the equation $i\sqrt{1-x^2}=-\sqrt{x^2-1}$ will only be true if $x\geq1$. In fact, if $x=1$, the equation is clearly true. If $0<x<1$, then $w=1-x^2>0$, so $i\sqrt{w}=-\sqrt{-w}=-i\sqrt{w}$, which is not true. Therefore, this identity holds for the interval $[1, \infty)$. Similar arguments can be constructed for other cases. – Emmanuel José García Feb 24 '24 at 14:16
  • @DavidK: We also have that $\tan{\frac12\sec^{-1}(x)}=\frac{\sqrt{x-1}}{\sqrt{x+1}}$. Rewriting the right side of identity $(1)$ in terms of these radicals also simplifies to $x-\sqrt{x^2-1}$. Therefore, this identity is also valid for the interval $[1, \infty)$ and not for $[0, 1)$, since for $x=0$ we get something like $i=-i$, which is not true. – Emmanuel José García Feb 24 '24 at 14:42
  • @DavidK: Consider again the equation $i\sqrt{1-x^2}=-\sqrt{x^2-1}$. For $x>1$, we have $w=1-x^2<0$, and therefore we have $i\sqrt{-w}=-\sqrt{w}$, which is true. – Emmanuel José García Feb 24 '24 at 15:09
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    I see you changed the conditions under which you use $(1)$ versus $(2)$, so now they agree with Wolfram Alpha. That resolves my previous comment, thank you. – David K Feb 24 '24 at 17:01
  • Mathematica gives: $-\frac{7 \sqrt{2}}{3}+4 \sqrt{3}-\sqrt{6}$ – David G. Stork Feb 25 '24 at 14:46
  • @DavidG.Stork: Yes, but can Mathematica solves this one? $$\int_{2}^{3} \frac{{1 - \tan\frac{{\sec^{-1}x}}{2}}}{{1 + \tan\frac{{\sec^{-1}x}}{2}}}\sqrt{\tan\frac{\csc^{-1}x}{2}},dx$$ I also posted my question on MO with some variations. The identities I have presented here definitely simplify some nearly impossible integrals. – Emmanuel José García Feb 25 '24 at 15:29
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    @EmmanuelJoséGarcía Let $x=\sec(2t)$ then yes it can. In particular, $$\int_{2}^{3}\frac{1-\tan\frac{\sec^{-1}x}{2}}{1+\tan\frac{\sec^{-1}x}{2}}\sqrt{\tan\frac{\csc^{-1}x}{2}},\mathrm{d}x=\frac{1}{5}\left(31\sqrt{2}-3\sqrt{6}-36\right).$$ – KStarGamer Feb 25 '24 at 15:46
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    After applying the substitution $x=\sec(2t)$, one arrives at $$I=2\int_{\frac{\pi}{6}}^{\frac{1}{2}\sec^{-1}3}\sec(2t)\tan(2t)\cdot\frac{1-\tan(t)}{1+\tan(t)}\sqrt{\cot\left(\frac{\pi}{4}+t\right)},\mathrm{d}t=-\frac{2}{5}\cdot\frac{2+3\sin(2t)}{1+\sin(2t)}\sqrt{\frac{\cos t-\sin t}{\cos t+\sin t}}\Bigg]_{\pi/6}^{\sec^{-1}(3)/2}.$$ – KStarGamer Feb 25 '24 at 15:51

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Too long for a comment:

@EmmanuelJoséGarcía. Yes Mathematica can solve your harder equation: $\left(\frac{1}{192}+\frac{i}{192}\right) \left(72 i \log \left(\text{Root}\left[\text{$\#$1}^{16}+8 \text{$\#$1}^{14}+44 \text{$\#$1}^{12}-136 \text{$\#$1}^{10}+230 \text{$\#$1}^8-136 \text{$\#$1}^6+44 \text{$\#$1}^4+8 \text{$\#$1}^2+1\&,7\right]\right)-(6-6 i) \sqrt[4]{3} \text{Root}\left[\text{$\#$1}^4+108\&,2\right] \log \left(\text{Root}\left[\text{$\#$1}^{16}+4 \text{$\#$1}^{14}+16 \text{$\#$1}^{12}-44 \text{$\#$1}^{10}+82 \text{$\#$1}^8-44 \text{$\#$1}^6+16 \text{$\#$1}^4+4 \text{$\#$1}^2+1\&,7\right]\right)-36 i \log \left(\text{Root}\left[\text{$\#$1}^{16}+4 \text{$\#$1}^{14}+16 \text{$\#$1}^{12}-44 \text{$\#$1}^{10}+82 \text{$\#$1}^8-44 \text{$\#$1}^6+16 \text{$\#$1}^4+4 \text{$\#$1}^2+1\&,7\right]\right)+36 i \log \left(\text{Root}\left[\text{$\#$1}^8+8 \text{$\#$1}^7+44 \text{$\#$1}^6-136 \text{$\#$1}^5+230 \text{$\#$1}^4-136 \text{$\#$1}^3+44 \text{$\#$1}^2+8 \text{$\#$1}+1\&,5\right]\right)-\sqrt{2} \sqrt[4]{3} \log \left(1+(1-i) \sqrt[4]{3}-i \sqrt{3}\right) \text{Root}\left[\text{$\#$1}^4-8748\&,3\right]-372 i 6^{3/4} \, _2F_1\left(-\frac{1}{4},-\frac{1}{4};\frac{3}{4};\frac{3}{2}\right)+942 i \sqrt[4]{2} \, _2F_1\left(-\frac{1}{4},-\frac{1}{4};\frac{3}{4};2\right)+744 i 6^{3/4} \, _2F_1\left(-\frac{1}{4},\frac{3}{4};\frac{7}{4};\frac{3}{2}\right)-2640 i \sqrt[4]{2} \, _2F_1\left(-\frac{1}{4},\frac{3}{4};\frac{7}{4};2\right)-8 i 6^{3/4} \, _2F_1\left(\frac{3}{4},\frac{3}{4};\frac{7}{4};\frac{3}{2}\right)+64 i \sqrt[4]{2} \, _2F_1\left(\frac{3}{4},\frac{3}{4};\frac{7}{4};2\right)+(-240+120 i) \sqrt[4]{2}-375\ 2^{3/4}+(72-72 i) \sqrt[4]{3}+(162-162 i) 3^{3/4}+15 i 2^{3/4} \sqrt{657+200 \sqrt{2}}-36 \log \left(1-(1-i) \sqrt[4]{2}-i \sqrt{2}\right)+36 \log \left(1+(1-i) \sqrt[4]{2}-i \sqrt{2}\right)-72 i \log \left(1-i \sqrt[4]{3}\right)+36 \log \left(1-(1-i) \sqrt[4]{3}-i \sqrt{3}\right)-(36-18 i) \log \left(1+(1-i) \sqrt[4]{3}-i \sqrt{3}\right)\right)$

David G. Stork
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