Problem 52 - Calculus Early Transcendentals

Question

Evaluate the following definite $$\int_{\sqrt{2}}^{2} \frac{\sqrt{x^{2}-1}}{x} d x$$

Step-by-Step Solution

Answer

Finish this statement: "The value of the definite integral $\int_{\sqrt{2}}^{2} \frac{\sqrt{x^{2}-1}}{x} dx$ is..." Answer: The value of the definite integral $\int_{\sqrt{2}}^{2} \frac{\sqrt{x^{2}-1}}{x} dx$ is $\frac{1}{2}\left(\operatorname{arcosh}(2)-\operatorname{arcosh}(\sqrt{2})+\frac{\sinh \left(2 \operatorname{arcosh}(\sqrt{2})\right)-\sinh \left(2 \operatorname{arcosh}(2)\right)}{2}\right)$.

Step 1: Identify the substitution

Let's make the substitution \(x = \cosh(u)\), because we have the expression \(\sqrt{x^2 - 1}\) in the integrand, and the identity \(\cosh^2(u) - 1 = \sinh^2(u)\) may come in handy. We'll also need to find \(dx\) in terms of \(du\): $$\frac{d x}{d u}=\frac{d}{d u}\cosh (u)=\sinh (u) \Rightarrow d x=\sinh (u) d u$$

Step 2: Change the limits of integration

Since we are changing the variable from \(x\) to \(u\), we also need to change the limits of integration correspondingly. For the lower limit, when \(x=\sqrt{2}\), let's find the corresponding value of \(u\): $$\sqrt{2}=\cosh (u) \Rightarrow u=\operatorname{arcosh}(\sqrt{2})$$ For the upper limit, when \(x=2\), we have: $$2=\cosh(u) \Rightarrow u=\operatorname{arcosh}(2)$$

Step 3: Rewrite the integral in terms of the new variable

Now let's rewrite the entire integral in terms of the new variable \(u\) and the new limits of integration: $$\int_{\sqrt{2}}^{2} \frac{\sqrt{x^{2}-1}}{x} dx = \int_{\operatorname{arcosh}(\sqrt{2})}^{\operatorname{arcosh}(2)} \frac{\sqrt{\cosh^2(u)-1}}{\cosh(u)}(\sinh(u)du)$$

Step 4: Simplify the integral

We can take advantage of the identity \(\cosh^2(u) - 1 = \sinh^2(u)\) to simplify the integrand: $$\int_{\operatorname{arcosh}(\sqrt{2})}^{\operatorname{arcosh}(2)} \frac{\sqrt{\sinh^2(u)}}{\cosh(u)}(\sinh(u)du) = \int_{\operatorname{arcosh}(\sqrt{2})}^{\operatorname{arcosh}(2)} \frac{\sinh(u)}{\cosh(u)}(\sinh(u)du)$$ $$\int_{\operatorname{arcosh}(\sqrt{2})}^{\operatorname{arcosh}(2)} \sinh^{2}(u)du$$

Step 5: Evaluate the integral

Now, let's integrate \(\sinh^2(u)\) with respect to \(u\) and apply the limits of integration: $$\int \sinh^2(u)du = \int \frac{1}{2}(1-\cosh(2u))du = \frac{1}{2}(u-\frac{\sinh(2u)}{2})+C$$ Now, apply the limits of integration: $$\int_{\operatorname{arcosh}(\sqrt{2})}^{\operatorname{arcosh}(2)} \sinh^{2}(u) d u=\frac{1}{2}\left(\operatorname{arcosh}(2)-\frac{\sinh \left(2 \operatorname{arcosh}(2)\right)}{2}\right) - \frac{1}{2}\left(\operatorname{arcosh}(\sqrt{2})-\frac{\sinh \left(2 \operatorname{arcosh}(\sqrt{2})\right)}{2}\right)$$

Step 6: Simplify the final result

Let's simplify the expression to find the final result: $$\frac{1}{2}\left(\operatorname{arcosh}(2)-\operatorname{arcosh}(\sqrt{2})+\frac{\sinh \left(2 \operatorname{arcosh}(\sqrt{2})\right)-\sinh \left(2 \operatorname{arcosh}(2)\right)}{2}\right)$$ So, the value of the definite integral is: $$\int_{\sqrt{2}}^{2} \frac{\sqrt{x^{2}-1}}{x} d x = \frac{1}{2}\left(\operatorname{arcosh}(2)-\operatorname{arcosh}(\sqrt{2})+\frac{\sinh \left(2 \operatorname{arcosh}(\sqrt{2})\right)-\sinh \left(2 \operatorname{arcosh}(2)\right)}{2}\right)$$

Key Concepts

Hyperbolic Functions

Hyperbolic functions are a set of functions that bear a resemblance to trigonometric functions, with the key difference being that they are defined using the exponential function. They are often applied to solving problems involving hyperbolas in the same way trigonometric functions are applied to circles.

Understanding the Basics

Hyperbolic sine and cosine, denoted as \(\text{sinh}(x)\) and \(\text{cosh}(x)\), are the two primary hyperbolic functions and are defined using the exponential function as follows:

• \(\text{sinh}(x) = \frac{e^x - e^{-x}}{2}\)
• \(\text{cosh}(x) = \frac{e^x + e^{-x}}{2}\)

These definitions allow us to derive various properties and identities, such as \(\text{cosh}^2(x) - \text{sinh}^2(x) = 1\), which mirrors the Pythagorean identity for trigonometric functions. The use of these functions can simplify the integration process when dealing with square roots involving a difference, just as we see in the provided exercise.

U-Substitution

U-substitution is a technique for evaluating integrals that can be seen as the reverse process of the chain rule used in differentiation. It simplifies integrals by transforming them into a more manageable form.

Applying U-Substitution

When faced with a complex integral, one can often use u-substitution to make it more straightforward. The process involves choosing a portion of the integral to substitute, which then simplifies the integral. As part of this, the differential \(dx\) is also substituted with \(du\), translating the entire integral into terms of the variable \(u\).

• Identify the substitution and let \(u\) equal a function of \(x\).
• Compute \(du\), the differential of \(u\) in terms of \(x\).
• Change the integral from \(x\) to \(u\), including the limits if it is a definite integral.

In our exercise, we substituted \(x\) with \(\cosh(u)\) and correspondingly altered \(dx\) to \(\sinh(u)du\), which made the integral more tractable.

Integration Limits

When performing a definite integral, the integration limits define the interval over which the function is integrated. These limits are critical when converting integration in terms of one variable to another, such as in u-substitution.

Changing Limits During U-Substitution

After performing u-substitution, the original limits, which are in terms of \(x\), must also be converted to the new variable \(u\). This is done by evaluating the substitution function at the original limits.

• Calculate the new lower limit by substituting the original lower limit into the substitution function.
• Calculate the new upper limit similarly.

These new limits ensure that the area under the curve defined by the integral remains the same despite the change of variable. In our exercise, the limits \(\sqrt{2}\) and \(2\) were converted to \(\operatorname{arcosh}(\sqrt{2})\) and \(\operatorname{arcosh}(2)\), after which the integral was evaluated.