Problem 88 - Thomas Calculus

Question

Use the integral tables to evaluate the integrals. \(\int \frac{\cosh ^{4} \sqrt{x}}{\sqrt{x}} d x\)

Step-by-Step Solution

Answer

\( \int \frac{\cosh ^{4} \sqrt{x}}{\sqrt{x}} \, dx = 2\left(\frac{3}{8}\sqrt{x} + \frac{1}{2}\sqrt{x} \sinh(2\sqrt{x}) + \frac{1}{32}\sinh(4\sqrt{x}) \right) + C \).

Step 1: Recognize the Form of the Integral

The given integral is \( \int \frac{\cosh^{4} \sqrt{x}}{\sqrt{x}} \, dx \). Notice that it involves the hyperbolic cosine function raised to the fourth power divided by the square root of \( x \). This hints at needing a substitution to simplify the integral.

Step 2: Use a Trigonometric Substitution

Let \( u = \sqrt{x} \), hence \( du = \frac{1}{2\sqrt{x}} \, dx \) which gives \( 2u \, du = dx \). Substitute into the integral:\[\int \cosh^{4}(u) \, 2u \, du \, / \, u^2 \ = \int \frac{2\cosh^{4}(u)}{u} \, du\]

Step 3: Refer to Integral Table

Look for a formula in the integral table that matches the form \( \int \cosh^{n}(u) \, du \). The integral table may list a result for \( n = 4 \). Use this formula to integrate \( \cosh^{4}(u) \).

Step 4: Integrate Using the Table Formula

The integral table provides:\[\int \cosh^{4}(u) \, du = \frac{3}{8}u + \frac{1}{2}u \cdot \sinh(2u) + \frac{1}{32}\sinh(4u) + C\]Multiply the above by 2 based on the substitution, resulting in:\[\int \frac{2\cosh^{4}(u)}{u} \, du = 2\left(\frac{3}{8}u + \frac{1}{2}u \sinh(2u) + \frac{1}{32}\sinh(4u) \right) + C\]

Step 5: Resubstitute in Terms of x

Reestablish \( u = \sqrt{x} \) to express the result in terms of \( x \). Substitute back:\[\int \frac{\cosh^{4} \sqrt{x}}{\sqrt{x}} \, dx = 2\left(\frac{3}{8}\sqrt{x} + \frac{1}{2}\sqrt{x} \sinh(2\sqrt{x}) + \frac{1}{32}\sinh(4\sqrt{x}) \right) + C\]

Key Concepts

Hyperbolic Functions

Hyperbolic functions are mathematical functions that share qualities with trigonometric functions but are based on hyperbolas rather than circles. They include hyperbolic sine, cosine, and tangent, denoted as \( \sinh(x) \), \( \cosh(x) \), and \( \tanh(x) \) respectively.
These functions arise naturally in areas such as the study of hyperbolic geometry, certain physics problems, and engineering applications. They can be defined using exponential functions:

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

Like their trigonometric counterparts, hyperbolic functions satisfy various identities, like \( \cosh^2(x) - \sinh^2(x) = 1 \), comparable to the Pythagorean identity for sine and cosine.
Understanding these functions is essential for working with many integrals, especially when they appear in exponential forms.

Trigonometric Substitution

Trigonometric substitution is a technique used in integral calculus to simplify integrals by substituting variables with trigonometric functions. This often allows for more manageable forms that align with known integral identities or formulas.
To use trigonometric substitution, you generally replace parts of the integral involving square roots or quadratic polynomials with trigonometric identities. In our case, substituting \( u = \sqrt{x} \) simplified the integral involving \( \cosh^{4}(u) \).

• For an integral involving \( \sqrt{a^2 - x^2} \), you might use \( x = a \sin(\theta) \).
• For \( \sqrt{a^2 + x^2} \), use \( x = a \tan(\theta) \).

This method is powerful as it can transform a difficult integral into a standard form that's easier to solve.

Integral Tables

Integral tables are valuable tools in calculus that provide formulas for integrals of various standard functions. These tables save time by allowing students to match their integral with a standard form and directly write the result without performing detailed calculations.
It's especially beneficial when dealing with complicated functions that are challenging to integrate directly, like \( \cosh^4(u) \) in our example.
The integral table used in the solution provided: \[ \int \cosh^{4}(u) \, du = \frac{3}{8}u + \frac{1}{2}u \cdot \sinh(2u) + \frac{1}{32}\sinh(4u) + C \] Using tables effectively requires identifying the correct form of your integral and knowing when it's appropriate to refer to them, often occurring after simplifying the expression with substitutions.

Definite and Indefinite Integrals

In calculus, integrals can be classified as either definite or indefinite. Indefinite integrals are used to find antiderivatives of a function, representing a family of functions whose derivative is the original function. They include a constant \(C\) because antiderivatives are not unique.
In contrast, definite integrals calculate the area under a curve between two points and provide a specific numerical value.
The integral \( \int \frac{\cosh^{4} \sqrt{x}}{\sqrt{x}} \, dx \) from our example is indefinite, as it finds the antiderivative of the function.

• Indefinite: \( \int f(x) \, dx \), generally written with \( + C \)
• Definite: \( \int_{a}^{b} f(x) \, dx \), where \(a\) and \(b\) are the limits of integration

Understanding the difference is crucial in solving calculus problems, as it impacts the approach and final result.