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Square roots with plus signs inside, like x2+9\sqrt{x^2 + 9}, often appear in physics when calculating gravitational potential or arc length along a parabola.
Standard trigonometric substitution turns this into sec⁡θ\sec\theta, but hyperbolic substitution simplifies the entire radical in a single step without messy secant integrals.
How does a hyperbolic function instantly eliminate the square root?

The Hyperbolic Identity

The fundamental hyperbolic identity is cosh⁡2u−sinh⁡2u=1\cosh^2 u - \sinh^2 u = 1. Rearranging it gives 1+sinh⁡2u=cosh⁡2u1 + \sinh^2 u = \cosh^2 u.
Because cosh⁡u\cosh u is strictly positive for all real uu, the square root cosh⁡2u\sqrt{\cosh^2 u} simplifies directly to cosh⁡u\cosh u with no absolute value signs needed.
Now that the square root collapses into a single term, what happens to the rest of the integral?

Step-by-Step Derivation

To evaluate I=∫1x2+a2 dxI = \int \frac{1}{\sqrt{x^2 + a^2}}\,dx for a positive constant aa, let x=asinh⁡ux = a\sinh u. Differentiating both sides gives dx=acosh⁡u dudx = a\cosh u\,du.
Substituting both pieces into the integral produces I=∫acosh⁡ua2sinh⁡2u+a2 du=∫acosh⁡uacosh⁡u du=∫1 duI = \int \frac{a\cosh u}{\sqrt{a^2\sinh^2 u + a^2}}\,du = \int \frac{a\cosh u}{a\cosh u}\,du = \int 1\,du.
Integrating 11 gives u+Cu + C. Since x=asinh⁡ux = a\sinh u, we have u=arsinh⁡(xa)u = \operatorname{arsinh}\left(\frac{x}{a}\right).