(cod: P-73-57-2) Trigonometric substitutions: Integrals involving the radical expressions \[ \sqrt{a^2 - x^2}, \sqrt{a^2 + x^2}, \text{ or } \sqrt{x^2-a^2},\] where \(a\) is a positive number, can often be transformed into simpler integrals through an appropriate trigonometric substitution. Indeed, consider the trigonometric identities below: \begin{eqnarray} \label{rel1} 1 - \sin^2\, \theta &=& \cos^2{\theta}, \\ \nonumber & & \\ \label{rel2} 1 + \tan^2{\theta} &=& \sec^2{\theta}, \\ \nonumber & & \\\label{rel3} \sec^2{\theta} - 1 &=& \tan^2{\theta}. \end{eqnarray} If the integral involves the radical \(\sqrt{a^2 - x^2}\), it may be convenient to use the substitution \begin{eqnarray*} x &=& a \, \sin\, \theta, \, \, \theta \in [-\pi/2,\pi/2], \\ dx &=& a \cos{\theta} \, d\theta, \end{eqnarray*} Since, by relation (\ref{rel1}), we have \begin{eqnarray*}\sqrt{a^2 - x^2} &=& \sqrt{a^2(1 - \sin^2\, {\theta})} \\ & & \\ &=& \sqrt{a^2 \cos^2{\theta}} = a \cos{\theta}. \end{eqnarray*} On the other hand, if the integral involves the radical \(\sqrt{a^2 + x^2}\), it may be convenient to use the substitution \begin{eqnarray*} x &=& a \, \tan{\theta}, \, \, \theta \in (-\pi/2,\pi/2), \\ dx &=& a \sec^2{\theta} \, d\theta, \end{eqnarray*} Since, by relation (\ref{rel2}), we have \begin{eqnarray*}\sqrt{a^2 + x^2} &=& \sqrt{a^2(1 + \tan^2\, {\theta})} \\ & & \\ &=& \sqrt{a^2 \sec^2{\theta}} = a \sec{\theta}. \end{eqnarray*} Finally, if the integral involves the radical \(\sqrt{x^2 - a^2}\), it may be convenient to use the substitution \begin{eqnarray*} x &=& a \, \sec{\theta}, \, \, \small{\theta \in (0,\pi/2) \text{ or } \theta \in (\pi,3\pi/2)}, \\ dx &=& \sec^2{\theta} \, d\theta, \end{eqnarray*} Since, by relation (\ref{rel3}), we have \begin{eqnarray*}\sqrt{x^2-a^2} &=& \sqrt{a^2(\sec^2{\theta} - 1)} \\ & & \\ &=& \sqrt{a^2 \tan^2{\theta}} = a \tan{\theta}. \end{eqnarray*} It is worth noting that these substitutions may not guarantee that the integral becomes simple, depending on the complete expression of the integrand. On the other hand, we observe that these substitutions can also be useful in cases where the root does not appear in the integral, that is, they are not exclusive to the cases addressed above, although they are more commonly used in this context.

Using an appropriate trigonometric substitution, we can ensure that the integral \[\int_{0}^{\frac{a\sqrt{2}}{2}} \dfrac{a}{\sqrt{a^2 - x^2}}\, dx\] where \(a\) is a positive number, is equal to: