Question

Determine the following indefinite integrals. \(\int \frac{d x}{\sqrt{x^{2}-16}}\)

Short answer

Question: Find the indefinite integral of the given function: \(\int \frac{1}{\sqrt{x^{2}-16}} dx\). Answer: \(\int \frac{1}{\sqrt{x^{2}-16}} dx = \cosh^{-1}\left(\frac{x}{4}\right) + C\).

Step-by-step solution

Identify the similar function to the integral

The function we are given, \(\frac{1}{\sqrt{x^{2}-16}}\), resembles the derivative of the inverse hyperbolic sine function. Recall that: \(\frac{d}{dx}(\sinh^{-1}(x)) = \frac{1}{\sqrt{x^2 + 1}}\). The main difference between our function and the derivative of the inverse hyperbolic sine function is the minus sign instead of the plus sign, and the constant of 16 instead of 1.

Apply substitution

To simplify the integral, perform a substitution using trigonometric identities. Let \(x = 4\cosh(t)\), so \(dx = 4\sinh(t) dt\). Then, our integral becomes: \(\int \frac{1}{\sqrt{(4\cosh(t))^2-16}} (4\sinh(t) dt) = \int \frac{4\sinh(t)}{\sqrt{16\cosh^2(t)-16}} dt\)

Simplify the expression

Factor out the constant from the square root: \(\int \frac{4\sinh(t)}{\sqrt{16(\cosh^2(t)-1)}} dt\) Notice that \(\cosh^2(t) - 1 = \sinh^2(t)\). Thus, our integral becomes: \(\int \frac{4\sinh(t)}{\sqrt{16\sinh^2(t)}} dt\) Now, simplify further: \(\int \frac{4\sinh(t)}{4 \times \sinh(t)} dt\) This simplifies to: \(\int dt\)

Integrate with respect to t

Integrating with respect to \(t\) gives us: \(I = \int dt = t + C\) Where \(I\) is the integral and \(C\) is the constant of integration.

Substitute back in terms of x

To express our answer in terms of the original variable \(x\), we need to find \(t\) in terms of \(x\). From our substitution, we have: \(x = 4\cosh(t)\) Divide by 4: \(\frac{x}{4} = \cosh(t)\) Now, apply the inverse hyperbolic cosine function: \(t = \cosh^{-1}\left(\frac{x}{4}\right)\) Finally, substitute this expression for \(t\) into our integral: \(I = \cosh^{-1}\left(\frac{x}{4}\right) + C\)

Write the final answer

The indefinite integral of the given function is: \(\int \frac{dx}{\sqrt{x^{2}-16}} = \cosh^{-1}\left(\frac{x}{4}\right) + C\)

Key concepts

Understanding Hyperbolic Functions

Hyperbolic functions are an extension of trigonometric functions, commonly used in calculus to handle integrals and derivatives involving exponential terms. The key hyperbolic functions include the hyperbolic sine (\( \sinh(x) \)), hyperbolic cosine (\( \cosh(x) \)), and their respective inverses.

These functions can be defined as follows:

• Hyperbolic sine: \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)
• Hyperbolic cosine: \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)
• Inverse hyperbolic sine: \( \sinh^{-1}(x) = \ln(x + \sqrt{x^2 + 1}) \)
• Inverse hyperbolic cosine: \( \cosh^{-1}(x) = \ln(x + \sqrt{x^2 - 1}) \)

A key identity that is often used is: \( \cosh^2(x) - \sinh^2(x) = 1 \). This identity allows for simplifying expressions and solving integrals as seen in the problem where the substitution of hyperbolic functions simplifies the expression into an easily integrable form.

Trigonometric Substitution Simplified

Trigonometric substitution is a technique in calculus used to simplify integrals that contain square roots. The method involves substituting a variable with a trigonometric function to transform the integral into a simpler form.

For functions of the type \( \sqrt{x^2 - a^2} \), hyperbolic substitution is often useful. This is because the properties of hyperbolic functions can help to transform the integrand into a function involving hyperbolic identities.

• When you see an integral of the form \( \sqrt{x^2 - a^2} \), consider using a hyperbolic substitution like \( x = a \cosh(t) \).
• This substitution helps to leverage the identity \( \cosh^2(x) - \sinh^2(x) = 1 \), which can simplify the integral.

In the example above, substituting \( x = 4 \cosh(t) \) led to a straightforward integral of \( \int dt \), which then allowed easy integration and back substitution for the variable transformation.

Fundamentals of Calculus in Integration

Calculus is fundamentally about understanding change and area. Its two main branches are differential calculus and integral calculus. While differential calculus focuses on rates of change and slopes, integral calculus deals with accumulation, including areas under curves.

In the realm of integration, indefinite integrals represent the antiderivative of a function. Unlike definite integrals, they do not evaluate to a specific numerical value but instead represent a family of functions \( F(x) \) such that \( F'(x) = f(x) \).

• Indefinite integrals involve constants of integration, typically denoted as \( C \).
• Common methods for solving indefinite integrals include substitution (as shown in the original problem) and integration by parts.
• Using transformations, such as trigonometric substitution, can simplify more complex functions to find their antiderivatives.

The original exercise showcases how integral calculus allows us to find an antiderivative and understand the behavior of the function beneath its mathematical expression.