Solving 'impossible' integrals with a new (?) trick

Question

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, Mathematica is unable to solve this integral (as confirmed in the comments):

$$\int_{2}^{3} \frac{{1 - \tan\frac{{\sec^{-1}x}}{2}}}{{1 + \tan\frac{{\sec^{-1}x}}{2}}}\sqrt{\tan\frac{\csc^{-1}x}{2}}\,dx\tag{7}$$

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):

$$\int_{2}^{3} \ln(x) \sqrt{\tan\left(\frac12\csc^{-1}x\right)} \, dx\tag{9}$$

$$\int_{2}^{3} \sin\left(\tan\frac12\csc^{-1}{x}\right) \, dx\tag{10}$$

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 (especially if you consider that sine and cosine functions can be expressed in terms of tangents of half angles.). 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.

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 MathSE.

This story will continue on my blog.

Answer 1

Not a full answer, but too long for a comment.

The OP's relations are related to the Gudermannian, see, e.g., https://mathworld.wolfram.com/Gudermannian.html, as $$ \operatorname{gd}(i \theta_1) = \theta_2 $$ for $\theta_k$ in \eqref{463459_2}, the other cases have similar identities. The Gudermannian itself is related to the limit $m\to1$ (or $k\to1$) of the parameter $m$/modulus $k$ of Jacobi elliptic functions, and is therefore sometimes called "hyperbolic amplitude", see also https://en.wikipedia.org/wiki/Gudermannian_function. Elliptic functions are known to fulfill a large number of identities, especially when combined with Jacobi’s Imaginary Transformation (https://mathworld.wolfram.com/JacobisImaginaryTransformation.html) of the complex plane. These identities give rise to many identities for the Gudermannian.

Answer 2

For the pleasure of working an interesting integral. $$I=\int\sqrt{\tan{\left(\frac{\csc^{-1}(x)}{2}\right)}}\,dx$$ $$x=\csc (2 t)\quad \implies \quad dx=-2 \cot (2 t) \csc (2 t)\,dt$$ $$I=-2 \int \cot (2 t) \csc (2 t) \sqrt{\tan (t)}\,dt$$ $$t=\tan ^{-1}\left(u^2\right)\quad \implies \quad dt=\frac{2 u}{u^4+1}\,du$$ $$I=\int \frac{u^4-1}{u^2}\,du=\frac{u^4+3}{3u}+C$$

Back to $x=\csc \left(2 \tan ^{-1}\left(u^2\right)\right)$

$$I=\frac{2}{3} \left(\sqrt{x^2-1}+2 x\right)\sqrt{x-\sqrt{x^2-1}}+C$$ For the definite integral $$\int_2^5\sqrt{\tan{\left(\frac{\csc^{-1}(x)}{2}\right)}}\,dx=4 \sqrt{3}-\sqrt{6}-\frac{7}{3} \sqrt{2}=1.17888$$

For the second integral $$J=\int\sqrt{e^{-i\sin^{-1}(x)}}\,dx$$ Mathematica returns a quite complicated result (have a look here). If the link is broken, just copy/paste in Wolfram Alpha

Integrate[Sqrt[E^(-I*ArcSin[x])],x]

You will need a lot of simplifications to obtain the result.

Edit

If, as @Carlo Beenakker did, we generalize the problem to $$I_p=\int \Bigg[\tan \left(\frac{1}{2} \csc ^{-1}(x)\right) \Bigg]^p\,dx$$

Using first $x=\csc (2 t)$ and then $t=\tan ^{-1}\left(u\right)$, we end with $$I_p=\frac{1}{2}\int \left(u^2-1\right) u^{p-2}\,du=\frac{1}{2} u^{p-1} \left(\frac{u^2}{p+1}+\frac{1}{1-p}\right)$$ where $$u=\tan \left(\frac{1}{2} \csc ^{-1}(x)\right)$$ that is to say (assuming $x>0$) $$I_p=\frac{\left(x-\sqrt{x^2-1}\right)^p \left(x+p\, \sqrt{x^2-1}\right)}{1-p^2}$$

Update

In comments, the poster asked for $$K=\int \frac{1-\tan \left(\frac{1}{2} \csc ^{-1}(x)\right)}{1+\tan \left(\frac{1}{2} \csc ^{-1}(x)\right)} \,\,\sqrt{\tan \left(\frac{1}{2} \csc ^{-1}(x)\right)} \,dx$$

Using the same sequence of changes of variables $$K=-\int \frac{\left(u^2-1\right)^2}{u^2}\,du=-\frac{u^3}{3}+2 u+\frac{1}{u}+C$$ which, back to $t$ is $$K=\frac{3+6 \tan (t)-\tan ^2(t)}{3 \sqrt{\tan (t)}}$$ Back to $x$ $$K=\frac{6 x \left(\sqrt{x^2-1}+x+1\right)+6 \sqrt{x^2-1}-4}{3 \left(\sqrt{x^2-1}+x\right)^{3/2}}$$