Evaluate the integral.
${\int }_{\sqrt{2}}^2\frac{1}{\mid x\mid \sqrt{x^2-1}}dx$

• A. $\frac{\pi }{12}$
• B. $-\frac{\pi }{12}$
• C. $\frac{7\pi }{12}$
• D. $\frac{5\pi }{12}$

Answer: (A)

Verified step by step guidance

1. Recognize that the integral involves an absolute value function, ∣x∣, and a square root term, √(x² - 1). This suggests that the integral may require a substitution to simplify the expression.
2. Since the integrand contains √(x² - 1), use the trigonometric substitution x = sec(θ), which simplifies √(x² - 1) to tan(θ). Recall that dx = sec(θ)tan(θ)dθ.
3. Substitute x = sec(θ) into the integral. The limits of integration will also change: when x = √2, θ = sec⁻¹(√2), and when x = 2, θ = sec⁻¹(2). The integral becomes ∫[sec⁻¹(√2)]^[sec⁻¹(2)] (1 / (|sec(θ)| * tan(θ))) * sec(θ)tan(θ)dθ.
4. Simplify the integrand after substitution. The |sec(θ)| term cancels with the sec(θ) in the numerator, and the tan(θ) terms also cancel, leaving ∫[sec⁻¹(√2)]^[sec⁻¹(2)] dθ.
5. Evaluate the integral ∫[sec⁻¹(√2)]^[sec⁻¹(2)] dθ, which is simply θ evaluated at the bounds. Substitute the limits of integration to find the result in terms of π.