Evaluate the integral. ∫ 2 2 1 ∣ x ∣ x 2 − 1 d x \int_{\sqrt2}^2\frac{1}{\left|x\right|\sqrt{x^2-1}}dx

we're asked to evaluate the integral from root two to two of one over the absolute value of x times the square root of x squared minus one d x. Now looking at this integral, it takes a very particular form that we can see results in an inverse secant. So this is equal to the inverse secant of X bounded from root two to two. So we can go ahead and just plug those bounds right in. So this is the inverse secant of two minus the inverse secant of the square root of two. Now, if you are not working with a calculator here or even if you are, it can be helpful to think of these inverse secants instead in terms of inverse cosines. So remember that if I have the inverse secant of something, I can rewrite it as the inverse cosine of the reciprocal of that. So I can think of this as the inverse cosine of one half minus the inverse cosine of one over root two, which is the same thing as root two over two. So root two over two. Then it's much easier to think about these in terms of our values on the unit circle here. When is the cosine equal to one half? It's when it's when our angle is Pi over three. So this is Pi over three minus the inverse cosine of root two over two. When is the cosine equal to root two over two? When that angle is Pi over four. So this is Pi over three minus Pi over four. Now if we want these to have a common denominator here, we can multiply pi over three by four over four and multiply pi over four by three over three to give them a common denominator of 12. So this becomes four pi minus three pi over 12, which ultimately is just one pi over 12, which we can write as pi over 12. So this gives us our final answer here to this definite integral. Feel free to let us know if you have any questions. I'll see you in the next one.

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Inverse Trigonometric Functions

Calculus

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Inverse Trigonometric Functions

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Multiple Choice

Evaluate the integral.
$\int_{\sqrt{2}}^{2} \frac{1}{∣ x ∣ \sqrt{x^{2} - 1}} d x$

$pi over 12$

$- \frac{π}{12}$

$7 pi over 12$

$5 pi over 12$

Verified step by step guidance

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.

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.

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.

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.

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.