- \(\displaystyle \int \frac {\sqrt {x^2-a^2}}{x} ~dx = \tan (\sec ^{-1} (\frac {x}{a})) - a \sec ^{-1} (\frac {x}{a}) = \color {blue} {\sqrt {x^2-a^2} - a \sec ^{-1} (\frac {x}{a})} = \color {blue} {\sqrt {x^2-a^2} - a \tan ^{-1} (\frac {\sqrt {x^2 - a^2}}{a})}\)
Solution: \(\sec (u) = \frac {1}{a} x\), \(\tan (u) \sec (u) ~du = \frac {1}{a} ~dx\)
Solution: \(\frac {\sqrt {x^2-a^2}}{x} ~dx = \frac {a \tan (u)}{a \sec (u)} \cdot a \tan (u) \sec (u) ~du = a \tan ^2 (u) ~du = a (\sec ^2 (u) - 1) ~du\)
\(\displaystyle \int \frac {\sqrt {x^2-1}}{x} ~dx = \color {blue} {\sqrt {x^2-1} - \sec ^{-1} (x)} = \color {blue} {\sqrt {x^2-1} - \tan ^{-1} (\sqrt {x^2-1})}\)
\(\displaystyle \int \frac {\sqrt {x^2-9}}{x} ~dx = \color {blue} {\sqrt {x^2-9} - 3 \sec ^{-1} (\frac {x}{3})} = \color {blue} {\sqrt {x^2-9} - 3 \tan ^{-1} (\frac {\sqrt {x^2-9}}{3})}\)
- Used 2019, Unsolved
Solution
Take \(a > 0\) and \(x > a\).
1. Substitute \(x = a \sec(\theta)\) with \(\theta \in (0, \pi/2)\).
1.1. Then \(dx = a \sec(\theta)\tan(\theta) \, d\theta\).
1.2. Since \(x > a > 0\), \(\sqrt{x^2 - a^2} = \sqrt{a^2(\sec^2(\theta) - 1)} = \sqrt{a^2 \tan^2(\theta)} = a \tan(\theta)\).
2. Transform and integrate: \[\int \frac{\sqrt{x^2-a^2}}{x} \, dx = \int \frac{a \tan(\theta)}{a \sec(\theta)} \cdot a \sec(\theta)\tan(\theta) \, d\theta = a \int \tan^2(\theta) \, d\theta.\]
2.1. From 1, \(x = a\sec(\theta)\), \(dx = a\sec(\theta)\tan(\theta)\,d\theta\), and \(\sqrt{x^2-a^2} = a\tan(\theta)\).
2.2. Substituting these into the integrand gives \(\frac{a\tan(\theta)}{a\sec(\theta)}\cdot a\sec(\theta)\tan(\theta)\,d\theta\).
2.3. The factors \(a\sec(\theta)\) cancel, leaving \(a\tan^2(\theta)\,d\theta\).
3. Evaluate \(a \int \tan^2(\theta) \, d\theta = a (\tan(\theta) - \theta) + C\).
Using the Pythagorean identity \(\tan^2(\theta) = \sec^2(\theta) - 1\): \[a \int (\sec^2(\theta) - 1) \, d\theta = a (\tan(\theta) - \theta) + C.\]
4. Express the antiderivative in terms of \(x\): \[\int \frac{\sqrt{x^2-a^2}}{x} \, dx = \sqrt{x^2-a^2} - a \operatorname{arcsec}\left(\frac{x}{a}\right) + C = \sqrt{x^2-a^2} - a \arctan\left(\frac{\sqrt{x^2-a^2}}{a}\right) + C.\]
4.1. Since \(\sec(\theta) = \frac{x}{a}\), we have \(\theta = \operatorname{arcsec}\left(\frac{x}{a}\right)\).
4.2. Since \(\tan(\theta) = \frac{\sqrt{x^2-a^2}}{a}\), we have \(a\tan(\theta) = \sqrt{x^2-a^2}\) and \(\theta = \arctan\left(\frac{\sqrt{x^2-a^2}}{a}\right)\).
4.3. Substituting these expressions into the antiderivative \(a(\tan(\theta) - \theta) + C\) from 3 gives the result.