- \(\displaystyle \int_{1}^{2} \frac {1}{x\sqrt {x^2 -1}} dx = \color {blue} {\frac {\pi}{3}}\)
- Solution: \(\sec ^{-1} (x) |_{1}^{2}\)
Solution
1. The antiderivative of \(\frac{1}{x\sqrt{x^2-1}}\) for \(x > 1\) is \(\operatorname{arcsec}(x) + C = \arccos(1/x) + C\).
1.1. Make the trigonometric substitution \(x = \sec(\theta)\) for \(\theta \in (0, \pi/2)\).
1.2. Then \(dx = \sec(\theta)\tan(\theta) \, d\theta\), and \(\sqrt{x^2 - 1} = \sqrt{\sec^2(\theta) - 1} = \tan(\theta)\).
1.3. Substituting: \[\int \frac{1}{x\sqrt{x^2 - 1}} \, dx = \int \frac{\sec(\theta)\tan(\theta)}{\sec(\theta)\tan(\theta)} \, d\theta = \int 1 \, d\theta = \theta + C = \operatorname{arcsec}(x) + C.\]
2. The definite integral is an improper integral at the lower limit \(x=1\), defined by \(\lim_{t \to 1^+} \int_t^2 \frac{1}{x\sqrt{x^2-1}} \, dx\).
The integrand has an infinite discontinuity at \(x=1\).
3. \[\int_1^2 \frac{1}{x\sqrt{x^2-1}} \, dx = \lim_{t \to 1^+} [\operatorname{arcsec}(x)]_t^2 = \operatorname{arcsec}(2) - \lim_{t \to 1^+} \operatorname{arcsec}(t) = \frac{\pi}{3} - 0 = \boxed{\frac{\pi}{3}}.\]
\(\sec(\pi/3) = 2\), so \(\operatorname{arcsec}(2) = \pi/3\); \(\sec(0) = 1\), so \(\operatorname{arcsec}(1) = 0\), and \(\operatorname{arcsec}\) is continuous at \(1\).