Problem 46 - Calculus of a Single Variable

Question

In Exercises 43–54, find the indefinite integral. $$ \int \frac{\cosh \sqrt{x}}{\sqrt{x}} d x $$

Step-by-Step Solution

Answer

The indefinite integral of \(\frac{\cosh\sqrt{x}}{\sqrt{x}}\) with respect to x is \(2\sinh(\sqrt{x}) + C\).

Step 1: Identify Suitable Substitution

Look at the integral \(\int \frac{\cosh \sqrt{x}}{\sqrt{x}} dx\). Notice that the derivative of \(\sqrt{x}\) is \(\frac{1}{2\sqrt{x}}\), which is similar to the denominator of the fraction. Hence, you can make the substitution \(u = \sqrt{x}\), giving \(du = \frac{1}{2\sqrt{x}} dx\) or \(2 du = \frac{dx}{\sqrt{x}}\).

Step 2: Substitute Variables

Substitute the new variables into the integral. In place of \(\sqrt{x}\) put \(u\), and replace \(dx\) with \(2 du\). This revision changes the integral to \(2\int \cosh(u) du\).

Step 3: Compute the Integral

The integral of the hyperbolic cosine function is the hyperbolic sine function. Hence, \(2\int \cosh(u) du = 2\sinh(u) + C\)

Step 4: Substitude back original variables

Finally, substitute back the original variables. In place of \(u\), put \(\sqrt{x}\). Your integral turns into \(2\sinh(\sqrt{x}) + C\).

Key Concepts

Substitution Method

The substitution method is a vital technique in calculus for simplifying integrals, especially when dealing with complex expressions. The main idea is to substitute a part of the integral with a new variable, often denoted by \( u \), making the integration easier to handle. In our exercise, we start by identifying suitable substitutions.

Here's how you can apply the substitution method:

• Look at the integral and identify where a substitution can simplify the expression. For example, if you see a composite function, it might be a candidate for substitution.
• Choose a part of the integral to substitute, usually by letting \( u \) equal some expression, then differentiate to find \( du \).
• Rewrite the entire integral in terms of \( u \). Make sure you replace all instances of the original variable.

A well-chosen substitution often transforms the integral into a simpler form where standard integration rules can be applied. Once the integral is evaluated, remember to substitute back the original variables to find the final answer.

Hyperbolic Functions

Hyperbolic functions, such as \( \cosh \) and \( \sinh \), are analogs of the trigonometric functions but pertain to hyperbolas instead of circles. In calculus, they appear in various contexts, including integration and differential equations. Understanding these functions and their properties is crucial for solving integrals involving them.

Here are some important points about hyperbolic functions:

• The hyperbolic cosine function \( \cosh(x) \) is defined as \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).
• The hyperbolic sine function \( \sinh(x) \) is defined as \( \sinh(x) = \frac{e^x - e^{-x}}{2} \).
• Just like their circular counterparts, \( \sinh \) and \( \cosh \) have well-known derivatives: \( \cosh'(x) = \sinh(x) \) and \( \sinh'(x) = \cosh(x) \).

In our exercise, understanding that the derivative of \( \cosh(x) \) is \( \sinh(x) \) simplifies the integration process once the substitution is made.

Integration Techniques

Integration techniques are the various methods used to find the indefinite integrals of functions. Different functions and combinations present unique challenges, and choosing the right method is key to simplifying the process. In our example, we used the substitution method, but there are other strategies as well.

A few common integration techniques include:

• Substitution: Helpful for integrals involving composite functions, as seen in our example.
• Integration by Parts: Useful for products of functions, stated as \( \int u dv = uv - \int v du \).
• Partial Fractions: Suitable for rational functions where the numerator's degree is less than the denominator's.
• Trigonometric Identities: Such as using identities to simplify integrands involving trigonometric functions.

In our exercise, using the substitution method was ideal because it transformed the integral into one involving a simple hyperbolic function, which was then straightforward to integrate.