Integration Using Some Euler-Like Identities

"Complexification formulas are great and it seems like this simplifies the right away."

- Ninad Munshi

 Some Euler-like identities

The following identities have been suggested based on formulas in this blog post.

Theorem 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. substitution

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:

\[\boxed{\int f\left(x,\tan{\frac{\beta}{2}}, \tan{\frac{\gamma}{2}} \right)\,dx=\int f\left(\frac{e^{i\alpha}+e^{-i\alpha}}{2}a, e^{\pm\text{i}\alpha}, \frac{1-e^{\pm\text{i}\alpha}}{1+e^{\pm\text{i}\alpha}}\right)\,\frac{e^{\mp\text{i}\alpha}-e^{\pm\text{i}\alpha}}{2i}a\,d\alpha}\tag{15}\]

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

$$\boxed{\int f\left(x, \sqrt{x^2-a^2}, \frac{\sqrt{x-a}}{\sqrt{x+a}}\right)\,dx= \int f\left(\frac{e^{i\alpha}+e^{-i\alpha}}{2}a, \frac{e^{\mp\text{i}\alpha}-e^{\pm\text{i}\alpha}}{2}a, \frac{1-e^{\pm\text{i}\alpha}}{1+e^{\pm\text{i}\alpha}}\right)\,\frac{e^{\mp\text{i}\alpha}-e^{\pm\text{i}\alpha}}{2i}a\,d\alpha}\tag{16}$$

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 comes from a question 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 7 illustrates how the technique presented in this blog turns out to be more effective by not requiring partial fraction decomposition.

Example 7. 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.

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.

Other integration techniques

• Integration by parts
• Integration by trigonometric substitution
• Weierstrass substitution
• Euler substitution
• Integration using Euler's formula
• Partial fraction decomposition
• Ostrogradski method
• Integration by reduction formulae
• Leibniz integral rule