Question

State the substitution you would make if you used trigonometric substitution and the integral involving the given radical, where \(a>0\). Explain your reasoning. (a) \(\sqrt{a^{2}-u^{2}}\) (b) \(\sqrt{a^{2}+u^{2}}\) (c) \(\sqrt{u^{2}-a^{2}}\)

Short answer

For \(\sqrt{a^{2}-u^{2}}\), substitute \(u=a\sin \theta\); for \(\sqrt{a^{2}+u^{2}}\), substitute \(u=a\tan \theta\); for \(\sqrt{u^{2}-a^{2}}\), substitute \(u=a\sec \theta\).

Step-by-step solution

Substitution for \(\sqrt{a^{2}-u^{2}}\)

Substitute \(u=a\sin \theta \). This is because \(\sin^2\theta\ + \cos^2\theta = 1 \) is one of the Pythagorean trigonometric identities, and rearranging this gives us \(\cos^2\theta = 1 - \sin^2\theta\), which is the form of our given expression. When we substitute \(u = a\sin\theta\), the expression \(\sqrt{a^{2}-u^{2}}\) becomes \(a\cos\theta\).

Substitution for \(\sqrt{a^{2}+u^{2}}\)

Substitute \(u=a\tan \theta\). This is because \(1 + \tan^2\theta = \sec^2\theta\) is another Pythagorean trigonometric identity, which is the form of our given expression. When we substitute \(u = a\tan\theta\), the expression \(\sqrt{a^{2}+u^{2}}\) becomes \(a\sec\theta\).

Substitution for \(\sqrt{u^{2}-a^{2}}\)

Substitute \(u=a\sec \theta \). This is because \(\sec^2\theta - 1 = \tan^2\theta\) is another Pythagorean trigonometric identity, which corresponds to having the \(u^2\) term positive while the \(a^2\) term is negative. When we substitute \(u = a\sec\theta\), the expression \(\sqrt{u^{2}-a^{2}}\) becomes \(a\tan\theta\).