Introduction
While investigating a way to further generalize my own
extension of Newton’s formula to arbitrary quadrilaterals, I stumbled—by sheer serendipity—upon an entire
family of trigonometric and hyperbolic formulas for the roots of quadratic equations. These formulas, which I originally showcased in a separate note, turned out to be surprisingly rich. In particular, they connect in elegant ways to certain identities useful for a variety of integrals. I have compiled the complete set of these quadratic root formulas (see the box of Formulas I–VI below), with further details posted in
A family of trigonometric formulas for the roots of quadratic equations.
Let me briefly highlight two classical techniques that intertwine in fascinating ways:
Euler’s celebrated identity, $e^{i\phi} = \cos(\phi) + i\sin(\phi)$, is well known for its power in simplifying integrals by converting sines and cosines into exponentials—and for enabling elegant treatments of Fourier transforms and other advanced tools in analysis.
Often referred to as the Weierstrass substitution, $t = \tan(\phi/2)$, this method simplifies numerous trigonometric integrals by transforming them into rational expressions. As Henning Dathe of the University of Göttingen details in
Die Halbwinkelsubstitution und ihre Anwendungen, half-angle substitutions find widespread use well beyond integral calculus, including stereographic projection, solving Riccati equations, trigonometric equations, rotation kinematics, and more. By the way, in an essay titled
The Theoretical Importance of Half-Angle Formulas, I have argued for the theoretical centrality of half-angle formulas in general.
Given this broad scope, one might wonder whether Euler’s formula and the tangent half-angle substitution are, in fact, just two sides of the same coin. Indeed, Theorem 1 below strongly suggests exactly that.
In the separate note alluded to above, I established several theorems (I–VI) for solving the quadratic equation
$$ax^2 + bx + c = 0\quad (a,b,c \neq 0),$$
each employing a different trigonometric or hyperbolic parameter. For clarity in this note, let us denote these parameters by $\alpha', \beta', \gamma', \delta', \eta', \theta'$ (instead of $\alpha, \beta, \gamma, \dots$) to avoid clashes with other symbols used here. The solutions can be gathered into the single box below:
$$\boxed{x_{1,2} \;=\;\begin{cases}\displaystyle\text{(I)} \quad i\,e^{\,i\alpha'}\!\sqrt{\tfrac{c}{a}},\;\;-\>i\,e^{\,i\alpha'}\!\sqrt{\tfrac{c}{a}}, & \alpha' = \sin^{-1}\!\Bigl(\tfrac{b}{2\sqrt{ac}}\Bigr),\\[6pt]
\displaystyle\text{(II)} \;-\>\tan\!\Bigl(\tfrac{\beta'}{2}\Bigr)\!\sqrt{\tfrac{c}{a}},\; -\>\cot\!\Bigl(\tfrac{\beta'}{2}\Bigr)\!\sqrt{\tfrac{c}{a}},& \beta' = \csc^{-1}\!\Bigl(\tfrac{b}{2\sqrt{ac}}\Bigr),\\[6pt]\displaystyle\text{(III)} \;\>e^{\,i\gamma'}\!\sqrt{\tfrac{c}{a}},\;\;-\>e^{-\,i\gamma'}\!\sqrt{\tfrac{c}{a}},& \gamma' = \cos^{-1}\!\Bigl(\tfrac{b}{2\sqrt{ac}}\Bigr),\\[6pt]\displaystyle
\text{(IV)} \;\;\frac{\tan\!\bigl(\tfrac{\delta'}{2}\bigr)\,\pm\,1}{\tan\!\bigl(\tfrac{\delta'}{2}\bigr)\,\mp\,1}\,\sqrt{\tfrac{c}{a}},
& \delta' = \sec^{-1}\!\Bigl(\tfrac{b}{2\sqrt{ac}}\Bigr),\\[6pt]
\displaystyle\text{(V)} \;\;\frac{i \,\pm\, e^{\eta'}}{\,i \,\mp\, e^{\eta'}}\,\sqrt{\tfrac{c}{a}},& \eta' = \tanh^{-1}\!\Bigl(\tfrac{b}{2\sqrt{ac}}\Bigr),\\[6pt]\displaystyle\text{(VI)} \;\;\frac{1 \,\pm\, e^{\theta'}}{\,1 \,\mp\, e^{\theta'}}\,\sqrt{\tfrac{c}{a}},
& \theta' = \coth^{-1}\!\Bigl(\tfrac{b}{2\sqrt{ac}}\Bigr).\\
\end{cases}}$$
In deriving these results, I noticed that Theorem 1 and Theorem 2 emerged naturally from equating and simplifying the respective expressions in (II)–(III) and in (III)–(IV). This interrelation, among others, underpins the unifying theme in these formulas and ties directly back to our original question of whether complex exponentials and half-angle substitutions are intimately related.
I am excited to present the
Unified Substitution Method (USM), a comprehensive approach that synthesizes and extends (to irrational integrals) two classical techniques for tackling integrals: the use of
complex exponentials for trigonometric functions and the
tangent half-angle substitution. In certain cases, the USM’s transformations (which are summarized at the end of this note) even yield rational expressions reminiscent of those derived through
Euler substitutions.
In what follows, you will see the journey I embarked upon in constructing this method. A
parallel paper, where
Theorem 1 and
Theorem 2 have been
generalized and
rigorously proven, is currently in development. Here, I present illustrative examples to showcase the method’s power. Enjoy!
Some Euler-like identitiesTheorem 1. If complex $\alpha=\cos^{-1}(x)$ and $\beta=\csc^{-1}(x)$, where real $x\in[1, \infty)$, then the following relation holds:
\[\boxed{e^{i\alpha}=\tan\left(\frac{\beta}{2}\right)}\tag{1}\]
In the complex plane, for real $x\in[0, 1]$, we have
\[\boxed{e^{-i\alpha}=\tan\left(\frac{\beta}{2}\right)}\tag{2}\]
Proof. Consider the left-hand side of equation $(1)$. By Euler's formula and properties of inverse trigonometric functions, the exponential $e^{i\cos^{-1}(x)}$ can be rewritten as follows.
\[e^{i\cos^{-1}{(x)}}=\cos{(\cos^{-1}{(x)})}+i\sin{(\cos^{-1}{(x)})}=x+i\sqrt{1-x^2}\]
The right-hand side of $(1)$ can be written as follows.
\[\tan{\left(\frac{\csc^{-1}{(x)}}{2}\right)}=\frac{\sin{(\csc^{-1}{(x)}})}{1+\cos{(\csc^{-1}{(x)}})}=\frac{\frac{1}{x}}{1+\frac{\sqrt{x^2-1}}{x}}=x-\sqrt{x^2-1}\]Now we just need to find the solution set for non-negative real $x$ of the following complex equation.
\[x+i\sqrt{1-x^2}=x-\sqrt{x^2-1}\tag{3}\]
Clearly, equation $(3)$ is true for $x=1$. For $x>1$, it follows that $1-x^2<0$, and therefore, the left-hand side of $(3)$ simplifies to $x-\sqrt{x^2-1}$. Consequently, equation $(3)$ is true for $x\geq1$.
For identity $(2)$, equation $(3)$ becomes
\[x-i\sqrt{1-x^2}=x-\sqrt{x^2-1}\tag{4}\]
If $0 \leq x \leq 1$, then $x^2 - 1 < 0$. Therefore, the right-hand side of equation $(4)$ can be rewritten as $x - i\sqrt{1 - x^2}$, making equation $(4)$ valid for $[0, 1]$ in the complex plane.
$\square$
The functions on both sides of identity $(1)$ are continuous for $x \geq 1$. And for identity $(2)$, continuity is guaranteed on $[0, 1]$ for complex numbers.
Remark. If complex $\alpha=\cos^{-1}(x)$ and $\beta=\csc^{-1}(x)$, where real $x\in[-1, 1]$, then the following relation holds:
\[\boxed{e^{i\alpha}=\cot\left(\frac{\beta}{2}\right)}\tag{5}\]
This is true because identity $(5)$ can be rewritten as:
\[x+i\sqrt{1-x^2}=x+\sqrt{x^2-1}\tag{6}\]
And the solution set of $(6)$ is $x \in [-1, 1]$.
Theorem 2. If complex $\alpha=\cos^{-1}(x)$ and $\gamma=\sec^{-1}(x)$, where real $x\in [1, \infty)$, then the following relation holds:
\[\boxed{e^{i\alpha}=\frac{1-\tan\left(\frac{\gamma}{2}\right) }{1+\tan\left(\frac{\gamma}{2}\right)}}\tag{7}\]In the complex plane, for $x\in [0, 1]$, we have
\[\boxed{e^{-i\alpha}=\frac{1-\tan\left(\frac{\gamma}{2}\right) }{1+\tan\left(\frac{\gamma}{2}\right)}} \tag{8}\]
Proof. By property of inverse trigonometric functions we have
\[\tan{\left(\frac{\sec^{-1}(x)}{2}\right)}=\frac{\sin{(\sec^{-1}(x))}}{1+\cos{(\sec^{-1}(x))}}=\frac{\frac{\sqrt{x^2-1}}{x}}{1+\frac{1}{x}}=\frac{\sqrt{x-1}}{\sqrt{x+1}}\]Substituting into the right-hand side of identity $(7)$, we obtain
$$\frac{1-\tan\left(\frac{\gamma}{2}\right)}{1+\tan\left(\frac{\gamma}{2}\right)}=\frac{1-\frac{\sqrt{x-1}}{\sqrt{x+1}}}{1+\frac{\sqrt{x-1}}{\sqrt{x+1}}}=x-\sqrt{x^2-1}$$
Hence, the same argument for Theorem 1 applies to Theorem 2, and both identities $(7)$ and $(8)$ are continuous on their respective intervals, with identity $(8)$ being valid in the complex plane.
$\square$
For $x\in[1,\infty)$, identity $(7)$ can be rewritten as follows:
\[\boxed{\tan{\left(\frac{\gamma}{2}\right)}=\frac{1-e^{i\alpha}}{1+e^{i\alpha}}}\tag{9}\]
And in the complex plane, for $x\in[0, 1]$, identity $(8)$ can be rewritten as follows:
\[\boxed{\tan{\left(\frac{\gamma}{2}\right)}=\frac{1-e^{-i\alpha}}{1+e^{-i\alpha}}}\tag{10}\]
Remark. For similar reasons to those given for identity $(5)$, for \(x \in [-1, 1]\), the following identity holds:
\[\boxed{\cot{\left(\frac{\gamma}{2}\right)}=\frac{1+e^{i\alpha}}{1-e^{i\alpha}}}\tag{11}\]
Applications
Example 1. Evaluate \[\int_{1}^{2}\sqrt[3]{\tan{\left(\frac12\csc^{-1}(x)\right)}} \, dx\tag{12}\]
Solution. Using identity $(1)$, integral $(12)$ becomes:
\[\int_{1}^{2} e^{\frac{1}{3}i\alpha}dx\]
Since $\alpha=\cos^{-1}(x)\implies d\alpha=-\frac{1}{\sqrt{1-x^2}}\,dx$ and $x=\cos{(\alpha)}$.
$$\begin{aligned}\int_{1}^{2} e^{\frac{1}{3}i\alpha}dx&= -\int_{1}^{2} e^{\frac{1}{3}i\alpha}\sqrt{1-\cos^2{(\alpha)}}\,d\alpha\\&=-\int_{1}^{2} e^{\frac{1}{3}i\alpha}\sin{(\alpha)}\,d\alpha\\&=-\frac{i}{2}\int_{1}^{2}(e^{-\frac23i\alpha}- e^{\frac43i\alpha})\,d\alpha\qquad (\text{Since $\sin{(\alpha)}=\frac12i(e^{-i\alpha}-e^{i\alpha}$}).)\\&=\frac34e^{-\frac23i\alpha}+\frac38e^{\frac43i\alpha}\bigg|_1^2\\&=\frac38e^{-\frac23i\alpha}(e^{2i\alpha}+2)\bigg|_1^2\\&=\frac{3\left((x-\sqrt{x^2-1})^2+2\right)}{8(x-\sqrt{x^2-1})^{2/3}}\bigg|_1^2\qquad (\text{Changing back to reals.})\\&=\frac{{3(2 + (2 - \sqrt{3})^2)}}{{8(2 - \sqrt{3})^{\frac{2}{3}}}} - \frac{9}{8}\\&\approx0.744\end{aligned}$$
Another approach involves using the simplification $\tan{\left(\frac12\csc^{-1}{(x)}\right)}=x-\sqrt{x^2-1}$. However, this method could lead to more complicated integrals, especially when evaluating rational trigonometric integrals with multiple terms involving $\tan{\left(\frac12\csc^{-1}{(x)}\right)}$ or $\tan{\left(\frac12\sec^{-1}{(x)}\right)}$ or both. Making an appropriate u-substitution might not be immediately obvious in those cases, whereas using identities $(1)$ and $(9)$ quickly turns the integral into an integral of a rational function. As an illustration, let's consider the following example.
Example 2. Evaluate
\[\int\frac{1}{\tan\left(\frac{1}{2}\csc^{-1} (x)\right) - \tan\left(\frac{1}{2}\sec^{-1}(x)\right)} \, dx\]
Solution. First, we rewrite the integral using the identity $(1)$ and $(9)$:
\[\int \frac{1}{e^{i\alpha} - \frac{1 - e^{i\alpha}}{1 + e^{i\alpha}}} \, dx=\int\frac{1+e^{i\alpha}}{e^{2i\alpha}+2e^{i\alpha}-1}\,dx\]Similar to how we proceeded in example 1, since $\alpha=\cos^{-1}(x)\implies d\alpha=-\frac{1}{\sqrt{1-x^2}}\,dx$ and $x=\cos{(\alpha)}$. Then
$$\begin{aligned}\int\frac{1+e^{i\alpha}}{e^{2i\alpha}+2e^{i\alpha}-1}\,dx &=-\int\frac{1+e^{i\alpha}}{e^{2i\alpha}+2e^{i\alpha}-1}\sqrt{1-\cos^2(\alpha)}\,d\alpha\\&=-\int\frac{1+e^{i\alpha}}{e^{2i\alpha}+2e^{i\alpha}-1}\sin(\alpha)\,d\alpha\\&=-\frac{i}{2}\int\frac{e^{-i\alpha} - e^{i\alpha} - e^{2i\alpha} + 1}{e^{2i\alpha}+2e^{i\alpha}-1}\,d\alpha\end{aligned}$$
Let $u=e^{i\alpha}\implies du=ie^{i\alpha}\,d\alpha$. Then, putting terms over a common denominator, and factoring, we have
$$-\frac{i}{2}\int\frac{e^{-i\alpha} - e^{i\alpha} - e^{2i\alpha} + 1}{e^{2i\alpha}+2e^{i\alpha}-1}\,d\alpha=\frac12\int\frac{(u - 1) (u + 1)^2}{u^2 (u^2 + 2 u - 1)}\,du$$
At this point, we can proceed by applying partial fraction decomposition.
Remark. In the MathSE forum, I asked about alternative approaches to the integral in example 2. Zacky has suggested the following substitution
$$\int \frac{1}{x-\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}}\,dx \overset{\frac{x-1}{x+1}=t^2}{=}4\int \frac{t}{(1+t)(1-t)^2(1-2t-t^2)}\,dt$$
However, as the reader may have noticed by the degree of the denominator, I would argue that this substitution leads us to a more complicated partial fraction decomposition. Look here and here and compare for yourself.
The following integral was asked by user SAQ on the MathSE forum.
Example 3. Evaluate
\[ \int\dfrac{\sqrt{x-1}-\sqrt{x+1}}{\sqrt{x-1}-3\sqrt{x+1}}\,dx\]
Solution. We first divide both the numerator and denominator by $\sqrt{x+1}$:
\[\int\dfrac{\sqrt{x-1}-\sqrt{x+1}}{\sqrt{x-1}-3\sqrt{x+1}}\,dx = \int\dfrac{\frac{\sqrt{x-1}}{\sqrt{x+1}}-1}{\frac{\sqrt{x-1}}{\sqrt{x+1}}-3}\,dx\]Now, we'll substitute $\frac{\sqrt{x-1}}{\sqrt{x+1}}$ with $\frac{1-e^{i\alpha}}{1+e^{i\alpha}}$ using the identity $(9)$:
\[= \int\dfrac{\frac{1-e^{i\alpha}}{1+e^{i\alpha}}-1}{\frac{1-e^{i\alpha}}{1+e^{i\alpha}}-3}\,dx\]As $\alpha = \cos^{-1}(x)$, then $x = \cos{\alpha}$ and $dx = -\sqrt{1-x^2}\,d\alpha = -\sqrt{1-\cos^2{(\alpha)}}\, d\alpha = -\sin{(\alpha)}\,d\alpha$. Now, we'll perform the substitution:
\[= -\int\dfrac{\frac{1-e^{i\alpha}}{1+e^{i\alpha}}-1}{\frac{1-e^{i\alpha}}{1+e^{i\alpha}}-3}\sin{(\alpha)}\,d\alpha\]Since $\sin(\alpha) = \frac{i}{2}(e^{-i\alpha} - e^{i\alpha})$ and simplifying, we get:
\[\frac{i}{2}\int \frac{e^{2i\alpha}-1}{2e^{i\alpha}+1} \, d\alpha\]Now, let $u = e^{i\alpha}$. Then, $du = ie^{i\alpha}\,d\alpha = iu\,d\alpha$ and $d\alpha = \frac{1}{iu}\,du$. Substituting, performing long division, and then decomposing into partial fractions,
\[ \begin{aligned}\frac12\int \frac{u^2 - 1}{u(2u + 1)} \, du&=\frac14\int 1\,du - \frac14\int \frac{u+2}{u(2u+1)}\,du\\&=\frac14\int1\,du -\frac12\int\frac{1}{u}\,du+\frac38\int\frac{1}{2u+1}\,du\\&=\frac{u}{4}-\frac{\ln{u}}{2}+\frac38\ln{|2u+1|}+C\\&=\frac{u}{4}+\frac12\ln{\left|\frac{(2u+1)^{\frac34}}{u}\right|}+C\\&=\frac{e^{i\alpha}}{4}+\frac12\ln{\left|\frac{(2e^{i\alpha}+1)^{\frac34}}{e^{i\alpha}}\right|}+C\\&=\frac{x-\sqrt{x^2-1}}{4}+\frac12\ln{\left|\frac{(2(x-\sqrt{x^2-1})+1)^{\frac34}}{x-\sqrt{x^2-1}}\right|}+C\end{aligned} \]
The following example has been taken from the popular YouTube channel, Blackpenredpen.
Example 4. Evaluate
$$\int\sqrt{\frac{1-x}{1+x}}\,dx$$
Solution. For $x\in[0, 1]$, applying identity $(10)$, we have
$$\begin{aligned}\int\sqrt{\frac{1-x}{1+x}}\,dx&=i\int\sqrt{\frac{x-1}{x+1}}\,dx\\&=-i\int\frac{1-e^{-i\alpha}}{1+e^{-i\alpha}}\sin(\alpha)\,d\alpha\\&=\frac12\int\frac{(e^{-i\alpha}-1)(e^{-i\alpha}-e^{i\alpha})}{1+e^{-i\alpha}}\,d\alpha\\&=\frac12\int((e^{-i\alpha}+e^{i\alpha})-2)\,d\alpha\\&=\frac12\int2(\cos(\alpha)-1)\,d\alpha\\&=\sin(\alpha)-\alpha+C\\&=\sqrt{1-x^2}-\cos^{-1}(x)+C\end{aligned}$$
You can arrive at this same expression by using the trigonometric substitution $x = \cos(t)$, as some commented on the channel.
An alternative method to trig./hyp./Euler substitutions
Perhaps the reader will have noticed that the technique described above serves as an alternative method to trigonometric substitution. In fact, if we adapt equations $(1)$ and $(9)$, we will have the following more general expressions:
$$e^{i\cos^{-1}\left(\frac{x}{a}\right)}=\tan\left(\frac{1}{2}\csc^{-1}\left(\frac{x}{a}\right)\right)=\frac{1}{a}(x-\sqrt{x^2-a^2})\tag{13}$$$$\frac{1-e^{i\cos^{-1}\left(\frac{x}{a}\right)}}{1+e^{i\cos^{-1}\left(\frac{x}{a}\right)}}=\tan\left(\frac{1}{2}\sec^{-1}\left(\frac{x}{a}\right)\right)=\frac{\sqrt{x-a}}{\sqrt{x+a}}\tag{14}$$
The general transformation formula is:
\[\boxed{\int f\left(x,\tan{\frac{\beta}{2}}, \tan{\frac{\gamma}{2}} \right)\,dx=\int f\left(\frac{e^{i\alpha}+e^{-i\alpha}}{2}a, e^{\pm\text{i}\alpha}, \frac{1-e^{\pm\text{i}\alpha}}{1+e^{\pm\text{i}\alpha}}\right)\,\frac{e^{-\text{i}\alpha}-e^{\text{i}\alpha}}{2i}a\,d\alpha}\tag{15}\]
Where $\alpha=\cos^{-1}\left(\frac{x}{a}\right)$, $\beta=\csc^{-1}\left(\frac{x}{a}\right)$ and $\gamma=\sec^{-1}\left(\frac{x}{a}\right).$
Also, you can use
$$\boxed{\int f\left(x, \sqrt{x^2-a^2}, \frac{\sqrt{x-a}}{\sqrt{x+a}}\right)\,dx= \int f\left(\frac{e^{i\alpha}+e^{-i\alpha}}{2}a, \frac{e^{\mp\text{i}\alpha}-e^{\pm\text{i}\alpha}}{2}a, \frac{1-e^{\pm\text{i}\alpha}}{1+e^{\pm\text{i}\alpha}}\right)\,\frac{e^{-\text{i}\alpha}-e^{\text{i}\alpha}}{2i}a\,d\alpha}\tag{16}$$
For the alternating signs $\mp$, use the upper sign when $\frac{x}{a} \geq 1$, and the lower sign when $0 \leq \frac{x}{a} \leq 1$.
The following example is a typical case where you would apply trigonometric substitution, specifically Case I, and it has been taken from the
Wikipedia page dedicated to this technique.
Example 5. Evaluate
$$\int_{-1}^{1}\sqrt{4-x^2}\,dx$$
Solution. Rewriting and applying transformation $(16)$, we have:
$$\begin{aligned}\int_{-1}^{1}\sqrt{4-x^2}\,dx&=i\int_{-1}^{1}\sqrt{x^2-4}\,dx\\&=\int_{-1}^{1}(e^{-i\alpha}-e^{i\alpha})^2\,dx\qquad (\text{Applying transformation $(16)$.})\\&=\int_{\frac{2\pi}{3}}^{\frac{\pi}{3}} \left(e^{-2i\alpha} + e^{2i\alpha} - 2\right) d\alpha\\&= \frac{1}{2}i e^{-2i\alpha} - \frac{1}{2}i e^{2i\alpha}-2\alpha\bigg|_\frac{2\pi}{3}^\frac{\pi}{3}\\&=\sin{(2\alpha)}-2\alpha\bigg|_\frac{2\pi}{3}^\frac{\pi}{3}\\&=\left( \sin \left( \frac{2\pi}{3} \right) - \frac{2\pi}{3} \right) - \left( \sin \left( \frac{4\pi}{3} \right) - \frac{4\pi}{3} \right)\\&=\sqrt{3} + \frac{2\pi}{3} \end{aligned}$$
This result coincides with the one obtained in the Wikipedia article. The
case III can also be tackled with this method. For
case II, we will make use of the following identity, valid for $x \in [0, \infty)$ and $a>0$:
$$e^{\sinh^{-1}{(\frac{x}{a})}}=\coth{\left(\frac12\text{csch}^{-1}{\left(\frac{x}{a}\right)}\right)}=\frac{1}{a}\left(\sqrt{x^2+a^2}+x\right)\tag{17}$$
Example 6. Evaluate
$$\int \sqrt{x^2+1}\,dx\tag{18}$$
Solution. From $(17)$ follows that
$$\int\sqrt{x^2+1}\,dx=\int e^{\sinh^{-1}{(x)}}\,dx-\int x\,dx$$
Now let $u=\sinh^{-1}(x)\implies dx=\sqrt{x^2+1}\,du$, and $\sinh^2{(u)}=x^2$. Then
$$\begin{aligned}\int e^{\sinh^{-1}{(x)}}\,dx&=\int e^{u}\sqrt{\sinh^2(u)+1}\,du\\&=\int e^{u}\cosh(u)\,du\\&=\frac12\int e^{u}\left(e^{-u}+e^{u}\right)\,du\\&=\frac12\int \left(1+e^{2u}\right)\,du\\&=\frac{u}{2}+\frac{e^{2u}}{4}+C\\&=\frac{\sinh^{-1}(x)}{2}+\frac{e^{2\sinh^{-1}(x)}}{4}+C \end{aligned}$$
Collecting back the terms,
$$\int\sqrt{x^2+1}\,dx=\int e^{\sinh^{-1}{(x)}}\,dx-\int x\,dx=\frac{2\sinh^{-1}(x)+e^{2\sinh^{-1}(x)}-2x^2}{4}+C$$
Compare this solution with the one provided in the Blackpenredpen and Integrals for you channels. The closed forms can be proven equivalent to the one given here easily; however, the solution given here seems much simpler to me.
The general transformation formula is:
$$\boxed{\int f\left(x, \sqrt{x^2+a^2}\right)\, dx = \int f\left(\frac{e^{\theta}-e^{-\theta}}{2}a, \frac{e^{\theta}+e^{-\theta}}{2}a\right) \frac{e^{\theta}+e^{-\theta}}{2}a\, d\theta}\tag{19}$$
Where $\theta=\sinh^{-1}(\frac{x}{a})$.
Applying $(19)$, the evaluation of integral $(18)$ is even more straightforward. The following example will make it clearer how effective $(19)$ can be.
Example 7 (2006 MIT Integration Bee). Evaluate
$$\int_{0}^{\infty} \frac{1}{\left(x+\sqrt{1+x^2}\right)^2}\,dx$$
Solution. Applying transformation $(19)$, the integral becomes
$$\begin{aligned}\int \frac{1}{\left(x+\sqrt{1+x^2}\right)^2}\,dx&=\frac12\int \frac{e^{\theta}+e^{-\theta}}{e^{2\theta}}\,d\theta\\&=\frac12\left[\int e^{-\theta}\,d\theta+\int e^{-3\theta}\,d\theta\right]\\&=\frac12\left[- e^{-\theta} - \frac{e^{-3\theta}}{3}+C\right]\\&=-\frac{1}{2\left(x + \sqrt{x^2 + 1}\right)} - \frac{1}{6\left(x + \sqrt{x^2 + 1}\right)^3}+C\qquad \left(\text{It follows from $(17)$.}\right)\\&=-\frac{{6x(\sqrt{x^2 + 1} + x) + 4}}{{6(\sqrt{x^2 + 1} + x)^3}}+C \end{aligned}$$
Now, applying the integration limits,
$$\int_{0}^{\infty} \frac{1}{\left(x+\sqrt{1+x^2}\right)^2}\,dx=-\frac{{6x(\sqrt{x^2 + 1} + x) + 4}}{{6(\sqrt{x^2 + 1} + x)^3}}\bigg|_0^\infty=\frac{2}{3}$$
In this
YouTube link, you can see how both competitors fail to evaluate this integral. The solution provided in this blog also appears less tedious than the one given by the
Let Solve Math Problems channel, using Euler substitution rather than the traditional trigonometric substitution.
A still more general version of $(19)$ is as follows:
$$\boxed{\int f\left(x, \sqrt{(x+b)^2+a^2}\right)\, dx = \int f\left(\frac{e^{\theta}-e^{-\theta}}{2}a-b, \frac{e^{\theta}+e^{-\theta}}{2}a\right) \frac{e^{\theta}+e^{-\theta}}{2}a\, d\theta}\tag{20}$$
Where $\theta=\sinh^{-1}(\frac{x+b}{a})$ and $a$ and $b$ are real numbers with $a>0$.
The following example also comes from a question at MathSE asked by user @user84413, where they complain that using the traditional substitutions $y=\sin(\beta)$ or $y=\tanh(\gamma)$ requires them to make a second substitution and then apply partial fraction decomposition. Example 8 illustrates how the technique presented in this blog turns out to be more effective by not requiring partial fraction decomposition.
Example 8. Evaluate
$$\int_0^1\frac{\sqrt{1-y^2}}{1+y^2}\,dy$$
Solution. By applying the following transformation formula
$$\int f\left(y, \sqrt{y^2-a^2}, \frac{\sqrt{y-a}}{\sqrt{y+a}}\right)\,dy= \int f\left(\frac{e^{-i\alpha}+e^{i\alpha}}{2}a, \frac{e^{i\alpha}-e^{-i\alpha}}{2}a, \frac{1-e^{-i\alpha}}{1+e^{-i\alpha}}\right)\,\frac{e^{i\alpha}-e^{-i\alpha}}{2i}a\,d\alpha,$$
where $\alpha=\cos^{-1}(y)$, $y\in[0, 1]$ and $a>0$, the integral becomes
$$\begin{aligned}\int \frac{\sqrt{1-y^2}}{1+y^2}\,dy &= \int \frac{{(e^{i \alpha} - e^{-i \alpha})^2}}{{(e^{i \alpha} - e^{-i \alpha})^2 + 8}}\, d\alpha \\&= \int \frac{\sin^2(\alpha)}{\sin^2(\alpha) - 2}\, d\alpha \\&= 2\int \frac{1}{\sin^2(\alpha) - 2}\,d\alpha + \int 1\,d\alpha\\&=\int1\,d\alpha-2\int \frac{\sec^2(\alpha)}{\tan^2(\alpha)+2}\,d\alpha\qquad \left(\text{Since $\sin(\alpha)=\frac{\tan(\alpha)}{\sec(\alpha)}$.}\right)\\&=\alpha-\sqrt{2}\tan^{-1}\left(\frac{\tan(\alpha)}{\sqrt{2}}\right)+C.\qquad \left(\text{Since $\int \frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\left(\frac{x}{a}\right) + C$.}\right) \end{aligned}$$
For the integration limits, we have that $\cos^{-1}(0)=\frac{\pi}{2}$ and $\cos^{-1}(1)=0$. Reversing the interval limits with a negative sign, we have
$$\begin{aligned}\int_{0}^{1} \frac{\sqrt{1-y^2}}{1+y^2}\,dy &= -\int_{0}^{\frac{\pi}{2}} \frac{\sin^2(\alpha)}{\sin^2(\alpha) - 2}\, d\alpha \\&= \sqrt{2} \tan^{-1}\left(\frac{\tan(\alpha)}{\sqrt{2}}\right) - \alpha\bigg|_{0}^{\frac{\pi}{2}}\\&=\lim_{{\alpha \to \frac{\pi}{2}^{-}}} \left( \sqrt{2} \tan^{-1}\left(\frac{\tan(\alpha)}{\sqrt{2}}\right) - \alpha \right)-0\\&=\frac{1}{2} (\sqrt{2}-1) \pi\end{aligned}$$
Alternatively, we have the option of converting the integral
$$\int \frac{{(e^{i \alpha} - e^{-i \alpha})^2}}{{(e^{i \alpha} - e^{-i \alpha})^2 + 8}}\, d\alpha$$
into an integral of a rational function and then applying partial fraction decomposition. However, this method, although more mechanical, would not be as simple as the solution we give above.
Example 9. Evaluate
$$\int x^2\sqrt{x^2-1}\,dx$$
Solution. After applying $(16)$ the integral becomes
$$\begin{aligned}\int x^2\sqrt{x^2-1}\,dx&=-\frac{i}{16}\int \left( e^{i\alpha} + e^{-i\alpha} \right)^2 \left( e^{-i\alpha} - e^{i\alpha} \right)^2 \, d\alpha\\&=-\frac{i}{16}\int \left(e^{-4 i \alpha} + e^{4 i \alpha} - 2 \right)\,d\alpha\\&=\frac{1}{64}e^{-4i\alpha} - \frac{1}{64} e^{4i\alpha} +\frac{i\alpha}{8} + C\\&=\frac{i}{32}(4\alpha - \sin(4\alpha)) + C\\&= \frac{i}{32}(4\cos^{-1}(x) - \sin(4\cos^{-1}(x))) + C\\&=\frac{1}{8}\left(\ln|x-\sqrt{x^2-1}| +x \left(2x^2-1\right) \sqrt{x^2-1}\right) + C \end{aligned}$$
Example 9 is a question by Chomowicz. I cordially invite the reader to compare the solution given in this blog with the solutions on MathSE.
$$\int \ln{\left(x+\sqrt{x^2-1}\right)}\,dx$$
Solution. Applying $(16)$ the integral becomes
$$\begin{aligned}\int\ln\left(x+\sqrt{x^2-1}\right)\,dx&=i\int \alpha\sin{\alpha}\,d\alpha\\&\overset{ibp}{=}i\left(-\alpha\cos{\alpha}+\int \cos{\alpha}\,d\alpha\right)\\&=i\left(\sin{\alpha}-\alpha\cos{\alpha}\right)+C\\&=i\left(\sin{\left(\cos^{-1}(x)\right)}-x\cos^{-1}(x)\right)+C\\&=i\sqrt{1-x^2}-x\color{red}{i\cos^{-1}(x)}+C\\&=-\sqrt{x^2-1}-x\color{red}{\ln{\left(x-\sqrt{x^2-1}\right)}}+C\\&=x\ln{\left(x+\sqrt{x^2-1}\right)}-\sqrt{x^2-1}+C \end{aligned}$$
Note: $\color{red}{i\cos^{-1}(x)=\ln{\left(x-\sqrt{x^2-1}\right)}}$ follows from identity $(1)$.
Integrals of the form $\int f\left(x,\frac{\sqrt{x+p}}{\sqrt{x+q}}\right)\,dx$
Let $b-a=p$ and $b+a=q$, where $a$ and $b$ are real numbers. Then the following identities hold:
$$\frac{1-e^{\pm\text{i}\alpha}}{1+e^{\pm\text{i}\alpha}}=\tan{\left(\frac12\sec^{-1}\left(\frac{x+b}{a}\right)\right)}=\frac{\sqrt{x+b-a}}{\sqrt{x+b+a}}=\frac{\sqrt{x+p}}{\sqrt{x+q}},\tag{21}$$
$$e^{\pm\text{i}\alpha}=\tan\left(\frac{1}{2} \csc^{-1}\left(\frac{x+b}{a}\right)\right) = \frac{x + b \mp \sqrt{(x + b)^2 - a^2}}{a}\tag{22}$$
These identities are generalizations of identities $(1-2)$ and $(9-10)$ and can be proven similarly. The general identities $(21)$ and $(22)$ lead us to the following general transformation formula:
$$\boxed{\int f\left(x,\tan{\frac{\beta}{2}}, \tan{\frac{\gamma}{2}} \right)\,dx=\int f\left(\frac{e^{i\alpha}+e^{-i\alpha}}{2}a-b, e^{\pm\text{i}\alpha}, \frac{1-e^{\pm\text{i}\alpha}}{1+e^{\pm\text{i}\alpha}}\right)\,\frac{e^{-\text{i}\alpha}-e^{\text{i}\alpha}}{2i}a\,d\alpha}\tag{23}$$
Or
$$\boxed{\int f\left(x, \sqrt{(x+b)^2-a^2}, \frac{\sqrt{x+b-a}}{\sqrt{x+b+a}}\right)\,dx= \int f\left(\frac{e^{i\alpha}+e^{-i\alpha}}{2}a-b, \frac{e^{\mp\text{i}\alpha}-e^{\pm\text{i}\alpha}}{2}a, \frac{1-e^{\pm\text{i}\alpha}}{1+e^{\pm\text{i}\alpha}}\right)\,\frac{e^{-\text{i}\alpha}-e^{\text{i}\alpha}}{2i}a\,d\alpha}\tag{24}$$
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 $0 \leq \frac{x + b}{a} \leq 1$.
Example 11. 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 $(23)$ 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, plug in the values of $a$ and $b$ into $(22)$ and simplify. Then, replace the expression in the antiderivative from $(22)$ and simplify again. Thus, we obtain
$$=\sqrt{x^2+3x+2} + \frac{1}{2} \ln\left|2x - 2\sqrt{x^2+3x+2} + 3\right|+C$$
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:
where as our method only requires a few lines of algebra and basic calculus, as in Example 11.
More generally, any integral of the form
$$\int \sqrt{\frac{\sqrt[n]{x} + p}{\sqrt[n]{x} + q}} \, dx,$$
where $n \in \mathbb{N}$, can be solved similarly as example 11, since after $u=x^{\frac{1}{n}}$ follows
$$\int \sqrt{\frac{u + p}{u + q}} \cdot n u^{n-1} \, du = n \int u^{n-1} \sqrt{\frac{u + p}{u + q}} \, du$$
Using the real and imaginary part of Euler's formula
Example 12. Evaluate
$$\int xe^{\cos^{-1}(x)}\,dx$$
Solution. Transforming the integral according to formula $(23)$ and using the real part of Euler's formula, we have
$$\begin{aligned}\int xe^{\cos^{-1}(x)}\,dx&=\text{Re}\left[-\frac{i}{2}\int e^{i\alpha}e^{\alpha}\left(e^{-i\alpha}-e^{i\alpha}\right)\,d\alpha\right]\\&=\text{Re}\left[-\frac{i}{2}\int \left(e^{\alpha}-e^{(1+2i)\alpha}\right)\,d\alpha\right]\\&=\text{Re}\left[-\frac{i}{2}\left(e^{\alpha}-\frac{e^{(1+2i)\alpha}}{1+2i}\right)+C\right]\\&=\frac{1}{10} e^{\alpha} \left( 2 \cos(2\alpha) - \sin(2\alpha) \right)+C\\&=\frac{1}{10} e^{\cos^{-1}(x)} \left( 2 \cos(2\cos^{-1}(x)) - \sin(2\cos^{-1}(x)) \right)+C\\&=-\frac{e^{\cos^{-1}(x)}}{5} \left(x \sqrt{1 - x^2} - 2x^2 + 1 \right)+C \end{aligned}$$
Disguised exponential substitution
Example 13. Evaluate
$$\int \sqrt{\sqrt{x-1}-\sqrt{x-2}} \, dx$$
The integral can be rewritten this way
$$\begin{aligned}\int \sqrt{\underbrace{\sqrt{x-1}-\sqrt{(\sqrt{x-1})^2-1}}_{e^{i\alpha}=f(x)-\sqrt{f(x)^2-1}}} \, dx \end{aligned}.$$
This suggest an exponential substitution since after
$$\cos{\alpha}=\sqrt{x-1},\qquad dx=-2\cos{\alpha}\sin{\alpha}\,d\alpha=-\sin{2\alpha}\,d\alpha=\frac{e^{-2i\alpha}-e^{2i\alpha}}{2i}\,d\alpha,$$
the integral becomes
$$\begin{aligned}\int \sqrt{\sqrt{x-1}-\sqrt{x-2}} \, dx&=-\frac{i}{2}\int e^{\frac12i\alpha}(e^{-2i\alpha}-e^{2i\alpha})\,d\alpha\\&=-\frac{i}{2}\int (e^{-\frac32i\alpha}-e^{\frac52i\alpha})\,d\alpha\\&=\frac{1}{3} e^{-\frac{3 i \alpha}{2}} + \frac{1}{5} e^{\frac{5 i \alpha}{2}} + C\\&=\frac13(\sqrt{x-1}+\sqrt{x-2})^{\frac32}+\frac15(\sqrt{x-1}-\sqrt{x-2})^{\frac52}+C \end{aligned}.$$
Example 14. Evaluate
$$\int \frac{1}{\sqrt{x}+\sqrt{x-1}+1} \, dx$$
The integral can be rewritten this way
$$\begin{aligned}\int \frac{1}{\underbrace{\sqrt{x}+\sqrt{(\sqrt{x})^2-1}}_{e^{-i\alpha}=f(x)+\sqrt{f(x)^2-1}} +1} \, dx \end{aligned}.$$
This suggest an exponential substitution since after
$$x=\cos^2{\alpha},\qquad dx=-2\cos{\alpha}\sin{\alpha}\,d\alpha=-\sin{2\alpha}\,d\alpha=\frac{e^{-2i\alpha}-e^{2i\alpha}}{2i}\,d\alpha,$$
the integral becomes
$$\begin{aligned}\int \frac{1}{\sqrt{x}+\sqrt{x-1}+1} \, dx&=-\frac{i}{2}\int\frac{e^{-2i\alpha}-e^{2i\alpha}}{e^{-i\alpha}+1}\,d\alpha\\&=-\frac{i}{2}\int\frac{e^{-i\alpha}-e^{3i\alpha}}{e^{i\alpha}+1}\,d\alpha\\&=\frac{i}{2}\int\frac{e^{-i\alpha}(e^{4i\alpha}-1)}{e^{i\alpha}+1}\,d\alpha\\&=\frac{i}{2}\int\frac{e^{-i\alpha}(e^{2i\alpha}+1)(e^{i\alpha}-1)(e^{i\alpha}+1)}{e^{i\alpha}+1}\,d\alpha\\&=\frac{i}{2}\int (e^{2i\alpha}-e^{i\alpha}-e^{-i\alpha}+1)\,d\alpha\\&=\frac12\left(\frac12e^{2i\alpha}+e^{-i\alpha}-e^{i\alpha}+i\alpha\right) + C\\&=\frac{1}{4}\left(2\ln\left|\sqrt{x} - \sqrt{x - 1}\right| - 2\sqrt{x - 1}\left(\sqrt{x} - 2\right) + 2x - 1 \right) + C \end{aligned}$$
Extended Weierstrass substitution
From $(22)$, by introducing the substitution $$t=e^{\pm\text{i}\alpha}=\tan\left(\frac{1}{2} \csc^{-1}\left(\frac{x+b}{a}\right)\right) = \frac{x + b \mp \sqrt{(x + b)^2 - a^2}}{a},$$ we can reformulate the transformation formula $(24)$ so that [extended] Weierstrass substitution can be applied directly to integrals with irrational integrands. This approach is neater than employing complex exponentials. However, on the interval $[-1, 1]$, using complex exponentials may be more convenient, since the tangent function becomes undefined at $\frac{x+b}{a}=0$.
With this substitution, we have:
$$\int f\left( x, \sqrt{(x + b)^2 - a^2}, \dfrac{ \sqrt{x + b - a} }{ \sqrt{x + b + a} } \right) \, dx= \int f\left(\frac{t^2+1}{2t}a - b, \mp\frac{t^2-1}{2t}a, \pm\frac{1-t}{1+t}\right) \frac{t^2-1}{2 t^2}a\, dt.\tag{25}$$
For the alternating signs $\mp$ and $\pm$:
- Use the upper sign when $\dfrac{x + b}{a} \geq 1$
- Use the lower sign when $\dfrac{x + b}{a} \leq -1$
Similarly, transformation formula $(20)$ can be rewritten as follows
$$\int f\left( x, \sqrt{(x + b)^2 + a^2}\right) \, dx= \int f\left(\frac{s^2-1}{2s}a - b, \frac{s^2+1}{2s}a\right) \frac{s^2+1}{2 s^2}a\, ds,\tag{26}$$
where $s=\coth{\left(\frac12\text{csch}^{-1}\left(\frac{x+b}{a}\right)\right)}=\frac{x+b+\sqrt{(x+b)^2+a^2}}{a}$, for $\frac{x+b}{a}>0$ and $a>0$.
Summary
Here are the general transformation formulas presented together:
I. Transformation for Integrals Involving $\tan\left( \frac12\csc^{-1}\left( \frac{x + b}{a} \right) \right)$ and $\tan\left( \frac12\sec^{-1}\left( \frac{x + b}{a} \right)\right)$:
$$\begin{aligned}\int f\biggl(x, \tan\left(\frac{\beta}{2}\right), \tan\left(\frac{\gamma}{2}\right)\biggr)\,dx&= \int f\left(\overbrace{\frac{e^{i\alpha} + e^{-i\alpha}}{2}}^{\cos \alpha} a - b,\; \overbrace{e^{\pm i\alpha}}^{\cos \alpha \pm i \sin \alpha},\; \overbrace{\frac{1 - e^{\pm i\alpha}}{1 + e^{\pm i\alpha}}}^{i \tan\left(\frac{\mp \alpha}{2}\right)}\right)\overbrace{\frac{e^{-i\alpha} - e^{i\alpha}}{2i}}^{-\sin \alpha} a \, d\alpha \\[6pt]&\stackrel{t=e^{\pm i\alpha}}{=} \int f\left(\frac{t^2+1}{2t}a - b,\; t,\; \pm\frac{1-t}{1+t}\right)\frac{t^2-1}{2t^2}a\, dt.\end{aligned}$$
- Where:
- $a>0$
- $\alpha = \cos^{-1}\left( \dfrac{x + b}{a} \right)$
- $\beta = \csc^{-1}\left( \dfrac{x + b}{a} \right)$
- $\gamma = \sec^{-1}\left( \dfrac{x + b}{a} \right)$
- For the alternating signs $\pm$:
- Use the upper sign when $\dfrac{x + b}{a} \geq 1$
- Use the lower sign when $\dfrac{x + b}{a} \leq -1$
- Use the lower sign when $0 \leq \dfrac{x + b}{a} \leq 1$ (valid in the complex plane)
II. Transformation for Integrals Involving $\sqrt{(x + b)^2 - a^2}$ and $\dfrac{\sqrt{x + b - a}}{\sqrt{x + b + a}}$:
$$\begin{aligned}\int f\Bigl( x, \sqrt{(x + b)^2 - a^2}, \frac{\sqrt{x + b - a}}{\sqrt{x + b + a}} \Bigr)\, dx&= \int f\Biggl(\overbrace{\frac{e^{i\alpha} + e^{-i\alpha}}{2}}^{\cos \alpha}a - b,\ \overbrace{\frac{e^{\mp i\alpha} - e^{\pm i\alpha}}{2}}^{\mp i \sin \alpha}a,\ \overbrace{\frac{1 - e^{\pm i\alpha}}{1 + e^{\pm i\alpha}}}^{i \tan\left(\frac{\mp \alpha}{2}\right)}\Biggr)
\overbrace{\frac{e^{-i\alpha} - e^{i\alpha}}{2i}}^{-\sin \alpha} a \, d\alpha \\[6pt]&\overset{t = e^{\pm i\alpha}}{=} \int f\Biggl(\frac{t^2 + 1}{2t}a - b,\ \mp \frac{t^2 - 1}{2t}a,\ \pm \frac{1-t}{1+t}\Biggr)\frac{t^2 - 1}{2t^2}a \, dt.\end{aligned}$$
- Where:
- $a>0$
- $\alpha = \cos^{-1}\left( \dfrac{x + b}{a} \right)$
- For the alternating signs $\mp$ and $\pm$:
- Use the upper sign when $\dfrac{x + b}{a} \geq 1$
- Use the lower sign when $\dfrac{x + b}{a} \leq -1$
- Use the lower sign when $0 \leq \dfrac{x + b}{a} \leq 1$ (valid in the complex plane)
III. Transformation for Integrals Involving $\sqrt{(x + b)^2 + a^2}$:
$$\begin{aligned}\int f\left( x, \sqrt{(x + b)^2 + a^2} \right)\, dx
&= \int f\Biggl(\overbrace{\frac{e^\theta - e^{-\theta}}{2}}^{\sinh \theta} a - b,\ \overbrace{\frac{e^\theta + e^{-\theta}}{2}}^{\cosh \theta} a\Biggr)\overbrace{\frac{e^\theta + e^{-\theta}}{2}}^{\cosh \theta} a\, d\theta \\[6pt]&\overset{s=e^\theta}{=} \int f\Biggl(\frac{s^2 - 1}{2s}a - b,\ \frac{s^2 + 1}{2s}a\Biggr)\frac{s^2 + 1}{2 s^2}a\, ds.\end{aligned}$$
- Where:
- $\theta = \sinh^{-1}\left( \dfrac{x + b}{a} \right)$
- $a > 0$
IV. Transformation for Integrals Involving $\tan\left( \frac12\sin^{-1}\left( \frac{x + b}{a} \right)\right)$ and $\tan\left( \frac12\cos^{-1}\left( \frac{x + b}{a} \right)\right)$:
$$\int f\left(x,\tan\left(\frac{1}{2}\psi\right), \tan\left(\frac{1}{2}\alpha\right) \right)\,dx=\int f\left(\frac{2r}{1+r^2}a-b, r, \pm\frac{1+r}{1-r}\right)\,\frac{2(1-r^2)}{(1+r^2)^2}a\,dr,$$
- Where:
- $a>0$
- $\phi = \sec^{-1}\left( \dfrac{x + b}{a} \right)$
- $\psi = \sin^{-1}\left( \dfrac{x + b}{a} \right)$
- $\alpha = \cos^{-1}\left( \dfrac{x + b}{a} \right)$- For the alternating signs $\pm$:
- Use the upper sign when $0\leq \dfrac{x+b}{a} \leq 1$
- Use the lower sign when $-1\leq \dfrac{x+b}{a} \leq 0$
- $r =e^{i\phi}=\tan\left(\frac{1}{2}\psi\right)=\dfrac{a-\sqrt{a^2-(x+b)^2}}{x+b}$
V. Transformation for Integrals Involving $\sqrt{a^2-(x+b)^2}$ and $\sqrt{\frac{a+b+x}{a-b-x}}$:
$$\int f\left(x, \sqrt{a^2-(x+b)^2}, \sqrt{\frac{a+b+x}{a-b-x}}\,\right)\,dx=\int f\left(\frac{2r}{1+r^2}a-b, \pm\frac{1-r^2}{1+r^2}a, \pm\frac{1+r}{1-r}\right)\,\frac{2(1-r^2)}{(1+r^2)^2}a\,dr,$$
- Where:
- $a>0$
- $\phi = \sec^{-1}\left( \dfrac{x + b}{a} \right)$
- $\psi = \sin^{-1}\left( \dfrac{x + b}{a} \right)$
- For the alternating signs $\pm$:
- Use the upper sign when $0\leq \dfrac{x+b}{a} \leq 1$
- Use the lower sign when $-1\leq \dfrac{x+b}{a} \leq 0$
- $r =e^{i\phi}=\tan\left(\frac{1}{2}\psi\right)=\dfrac{a-\sqrt{a^2-(x+b)^2}}{x+b}$
Remark: Very efficient for integrals of the form $\frac{1}{x^{n}\sqrt{a^2-(x+b)^2}}$ and $\frac{1}{x^{n}}\sqrt{\frac{a+b+x}{a-b-x}}$ since it reduce them to polynomial-like forms.
Regarding the originality of this technique, I've asked on Mathoverflow under the title 'Solving 'impossible' integrals with a new (?) trick.' Take a look for yourself.
