Problem 56 - Calculus Early Transcendentals

Question

Determine the following indefinite integrals. $$\int \frac{d x}{x \sqrt{16+x^{2}}}$$

Step-by-Step Solution

Answer

Question: Compute the indefinite integral of the function \(f(x) = \frac{1}{x\sqrt{16+x^2}}\). Answer: The indefinite integral of the function is given by \(\int f(x) dx = \frac{1}{4}\operatorname{arsinh}\left(\frac{x}{4}\right) + C\).

Step 1: Identify the proper substitution

Observe that the integral involves \(\sqrt{16+x^2}\). A standard substitution is to use \(x = 4\sinh u\) because then, \(dx = 4\cosh u du\). Performing this substitution would simplify the square root.

Step 2: Perform the substitution

Now, substitute \(x = 4\sinh u\). Thus, \(dx = 4 \cosh u du\). The integral becomes: $$\int \frac{d x}{x \sqrt{16+x^{2}}} = \int \frac{4 \cosh u du}{4\sinh u \sqrt{16+16\sinh^2 u}}$$

Step 3: Simplify the integral

Next, we simplify the integral. Notice that \(16 + 16\sinh^2 u = 16(\cosh^2 u)\) as \(\cosh^2 u-\sinh^2 u = 1\). Thus, the integral becomes: $$\int \frac{4 \cosh u du}{4\sinh u \sqrt{16(\cosh^2 u)}} = \int \frac{\cosh u du}{\sinh u 4\cosh u} = \frac{1}{4} \int \frac{du}{\sinh u}$$

Step 4: Use the definition of \(\operatorname{arsinh}\)

The remaining integral is well-known. It is the definition of the inverse hyperbolic sine function, which is often written as \(\operatorname{arsinh}\) or \(\operatorname{asinh}\): $$\frac{1}{4} \int \frac{du}{\sinh u} = \frac{1}{4} \operatorname{arsinh}(u) + C$$

Step 5: Substitute back and find the final result

Now, substitute back the original variable \(x\) to get the answer in terms of \(x\). Recall that \(x = 4\sinh u\), so we have \(u= \operatorname{arsinh} \left(\frac{x}{4}\right)\). Therefore, $$\frac{1}{4} \operatorname{arsinh}(u) + C = \frac{1}{4} \operatorname{arsinh}\left(\frac{x}{4}\right) + C$$ The indefinite integral is: $$\int \frac{d x}{x \sqrt{16+x^{2}}} = \frac{1}{4}\operatorname{arsinh}\left(\frac{x}{4}\right) + C$$

Key Concepts

Substitution Method

The substitution method is a powerful technique in calculus for simplifying integrals. It involves changing the variable of integration to rewrite an integral in a simpler form. This is particularly helpful when dealing with complicated expressions, enabling us to transform them into more manageable functions.

To apply the substitution method, we follow these steps:

• Identify a part of the integral that matches a substitution formula or recognizes a pattern similar to basic derivatives or integrals.
• Choose a substitution, such as setting a part of the integral to a new variable (e.g., let \(x = g(t)\)).
• Rewrite the differential (\(dx\)) in terms of the new variable \( (dt)\).
• Replace both the variable and differential in the integral to transform it into a function of the new variable.
• Simplify and solve the integral.
• Finally, substitute back the original variable to express your result in terms of the initial variable.

In our example, we used \(x = 4\sinh u\), which simplifies the expression under the square root, making the integral easier to solve. The substitution can significantly reduce complexity, transforming a challenging problem into a straightforward one.

Hyperbolic Functions

Hyperbolic functions are analogs to trigonometric functions but are related to the hyperbola, just as trigonometric functions are related to the circle. The most common hyperbolic functions include the hyperbolic sine \(\sinh(x)\) and the hyperbolic cosine \(\cosh(x)\). These functions are defined as:

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

Hyperbolic functions have properties similar to trigonometric functions, like periods and identities. For instance, the identity \(\cosh^2(x) - \sinh^2(x) = 1\) is akin to the Pythagorean identity \(\cos^2(\theta) + \sin^2(\theta) = 1\).

They are useful in solving problems involving certain types of integrals, especially when dealing with squares in expressions, as observed in our exercise. Transforming the integral using a hyperbolic function substitution helped simplify and solve the integral more efficiently, by taking advantage of these natural identities.

Inverse Hyperbolic Functions

Inverse hyperbolic functions are the inverses of the hyperbolic functions, and they are crucial in integration involving hyperbolic expressions. Some common inverse hyperbolic functions are \(\operatorname{arsinh}(x)\), \(\operatorname{arcosh}(x)\), and \(\operatorname{artanh}(x)\).

The inverse hyperbolic sine is given by:

• \(\operatorname{arsinh}(x) = \ln(x + \sqrt{x^2 + 1})\)

In our problem, we reach a stage where the integral \( \int \frac{du}{\sinh u} \) corresponds directly with the inverse hyperbolic sine function. Recognizing this correspondence allows us to express the result in terms of \(\operatorname{arsinh}(u)\), simplifying our integral to a concise expression. Finally, substituting back to the original variable provides a neat solution in terms of variables familiar from the initial problem statement.

Understanding and recognizing inverse hyperbolic functions in integrals is a powerful skill, extending your toolkit for tackling complex integration problems.