Emmanuel José García
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The half-angle formulas are central!

And as a picture is worth a thousand words...

Integration via Exponential Substitution

We introduce Exponential Substitution, which is a transformation that simplifies certain composite trigonometric integrals and offers an alternative approach to integrating through trigonometric, hyperbolic, and Euler substitutions. For a more detailed description of how this technique works, visit the blog post 'Integration Using Some Euler-like Identities.' Examples 2, 3, and 7 illustrate how this method could be more efficient than the traditional approaches.

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

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

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

For the alternating signs $$\mp$$, use the upper sign when $$\frac{x}{a} \geq 1$$, and the lower sign when $$0 \leq \frac{x}{a} \leq 1$$.

$$\boxed{\int f\left(x, \sqrt{x^2+a^2}\right)\, dx = \int f\left(\frac{e^{\theta}-e^{-\theta}}{2}a, \frac{e^{\theta}+e^{-\theta}}{2}a\right) \frac{e^{\theta}+e^{-\theta}}{2}a\, d\theta}\tag{3}$$

Where $$\theta=\sinh^{-1}(\frac{x}{a})$$ and $$a>0$$.

Although closely related to integration using Euler's formula, it is not exactly the same.

Discovery is the privilege of the child: the child who has no fear of being once again wrong, of looking like an idiot, of not being serious, of not doing things like everyone else. - A. Grothendieck