Question

In the following exercises, use the substitutions x=sinh?,cosh?, or tanh?. Express the final answers in terms of the variable x. $$ \int \frac{\sqrt{x^{2}-1}}{x^{2}} d x $$

Short answer

\( \int \frac{\sqrt{x^2 - 1}}{x^2} \, dx = \text{arccosh}(x) - \frac{\sqrt{x^2 - 1}}{x} + C \).

Step-by-step solution

Choose the Right Substitution

To solve the integral \( \int \frac{\sqrt{x^2 - 1}}{x^2} \, dx \), we'll use the substitution appropriate for handling expressions involving \( \sqrt{x^2 - 1} \). The hyperbolic identity \( x = \cosh(\theta) \) fits well, as \( \sinh^2(\theta) = \cosh^2(\theta) - 1 \).

Substitute and Derive Differential

Substitute \( x = \cosh(\theta) \), then \( dx = \sinh(\theta) \, d\theta \). Also, replace \( \sqrt{x^2 - 1} \) with \( \sinh(\theta) \) since \( \sqrt{\cosh^2(\theta) - 1} = \sinh(\theta) \).

Simplify the Integral

The integral transforms to \( \int \frac{\sinh(\theta)}{\cosh^2(\theta)} \sinh(\theta) \, d\theta = \int \frac{\sinh^2(\theta)}{\cosh^2(\theta)} \, d\theta \).

Use Trigonometric Identity

Recognize \( \sinh^2(\theta) = \cosh^2(\theta) - 1 \), so the integral becomes \( \int \frac{\cosh^2(\theta) - 1}{\cosh^2(\theta)} \, d\theta = \int \left( 1 - \frac{1}{\cosh^2(\theta)} \right) \, d\theta \).

Simplify Further

This simplifies to \( \int 1 \, d\theta - \int \text{sech}^2(\theta) \, d\theta = \theta - \tanh(\theta) + C \).

Return to Original Variable

Since \( \theta = \text{arcsech}(x) \), rewrite the result in terms of \( x \). We have \( \theta = \text{arccosh}(x) \) and \( \tanh(\theta) = \frac{\sinh(\theta)}{\cosh(\theta)} = \sqrt{x^2 - 1}/x \). Thus, the result is \[ \theta - \tanh(\theta) = \text{arccosh}(x) - \frac{\sqrt{x^2 - 1}}{x} + C \].

Key concepts

Integration Techniques

Integration techniques are methods used to find the integral of a function. They are crucial for solving problems involving areas under curves, among other applications. One important integration technique involves using substitutions. This method often simplifies complex integrals by changing variables to make the integration process more straightforward.

When using substitutions, you replace the original variable with a new one to simplify the mathematical expression. For example, in the given exercise, the expression inside the square root is simplified by a hyperbolic substitution. This leads to expressing the integral in a way that's easier to integrate using known identities.

Some common substitution techniques include trigonometric substitution, hyperbolic substitution, and polynomial substitution.

Hyperbolic Functions

Hyperbolic functions, such as sinh, cosh, and tanh, are mathematical functions that share similar properties to the trigonometric functions but are based on hyperbolas rather than circles.

Some common identities that involve hyperbolic functions include:

• \( ext{sinh}^2( heta) + 1 = ext{cosh}^2( heta) \)
• \( ext{cosh}^2( heta) - ext{sinh}^2( heta) = 1 \)
• \( ext{tanh}( heta) = rac{ ext{sinh}( heta)}{ ext{cosh}( heta)} \)

Understanding these functions and identities is crucial when solving integrals that involve expressions like \( ext{sqrt}(x^2 - 1) \). In the exercise, using \( ext{cosh} \) and \( ext{sinh} \) simplified the expression and made integration more straightforward. By substituting \( x = \text{cosh}( heta) \), the given integral becomes manageable and directly solvable with these identities.

Variable Substitution in Calculus

Variable substitution is a widely-used technique in calculus to simplify integrals and make them solvable. When an integral appears complex, substituting the original variable with a trigonometric or hyperbolic expression can often simplify the process.

In the given problem, the substitution \( x = \text{cosh}( heta) \) was chosen because it works well with the term \( \sqrt{x^2 - 1} \). By substituting, you can express this term as \( \text{sinh}( heta) \), which has much more straightforward integral properties. The entire integral is then reexpressed in terms of \( \theta \), which can be integrated using standard rules.

The replacement not only simplifies algebraic manipulation but also taps into useful identities, leading to easier integration steps. After integrating with respect to the new variable, it's important to substitute back to the original variable to complete the problem. This step allows expressing the final result in terms of the initial variable, providing a complete solution to the original problem.