Problem 42 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(41-50\) . $$ \int \sinh \frac{x}{5} d x $$

Step-by-Step Solution

Answer

\( \int \sinh \frac{x}{5} \, dx = 5 \cosh \left( \frac{x}{5} \right) + C \)

Step 1: Identify the antiderivative formula for hyperbolic sine

The integral of the hyperbolic sine function, \( \sinh(u) \), is \( \cosh(u) \) plus a constant of integration. Therefore, we need to find the antiderivative of \( \sinh \frac{x}{5} \).

Step 2: Simplify the Integral

Set \( u = \frac{x}{5} \), then \( du = \frac{1}{5} dx \). This means that \( dx = 5 \, du \). Substitute these into the integral to rewrite it as \( \int \sinh(u) \cdot 5 \, du \).

Step 3: Perform the Integration

Integrate \( 5 \int \sinh(u) \, du = 5 \cosh(u) + C \), where \( C \) is the constant of integration.

Step 4: Substitute Back for Original Variable

Replace \( u \) with \( \frac{x}{5} \) to get the antiderivative in terms of \( x \). This yields \( 5 \cosh \left( \frac{x}{5} \right) + C \).

Key Concepts

Antiderivative

The concept of the antiderivative is crucial in integral calculus. An antiderivative of a function is another function whose derivative matches the original function. To put it simply, it's like reversing the process of differentiation. By finding an antiderivative, we uncover a function whose rate of change with respect to the variable is given by the original function.
For example, the antiderivative of the hyperbolic sine function, \( \sinh(u) \), is the hyperbolic cosine function, \( \cosh(u) \). This means that if we differentiate \( \cosh(u) \), we get back to \( \sinh(u) \).
Finding antiderivatives can initially seem puzzling, but it follows certain rules and formulas. Antiderivatives are just one component of an indefinite integral, which expresses the entire family of functions, including a constant of integration, \( C \). This constant is crucial as any derivative of a constant is zero, indicating that multiple functions can have the same derivative.

Hyperbolic Functions

Hyperbolic functions, much like trigonometric functions, arise from exponential functions. They are especially useful in various fields such as calculus, physics, and engineering.
There are several key hyperbolic functions, but for this problem, we're focusing on \( \sinh(x) \) and \( \cosh(x) \). These functions are defined as:

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

These definitions mirror the relations found in trigonometry, such as sine and cosine. A unique property of hyperbolic functions is that they are linked to the geometry of hyperbolas.
The hyperbolic sine, \( \sinh \), describes how some values fall on a hyperbola, while \( \cosh \) handles the associated hyperbolic angle measures. These functions follow several rules similar to the familiar trigonometric identities, like \( \cosh^2(x) - \sinh^2(x) = 1 \), akin to the Pythagorean identity.

Integration Techniques

Integrating functions involves various techniques, and choosing the right one can simplify the process. When dealing with functions like \( \sinh \frac{x}{5} \), substitution is a valuable technique to simplify the integral.

• **Substitution Method:** This method changes variables to simplify the integral. For the given integral, we set \( u = \frac{x}{5} \), which leads to \( du = \frac{1}{5} dx \). Adjust the integral by substituting these values.
• **Identify Formulas:** Recognize standard integrals and derivatives, especially for hyperbolic functions. Knowing these can make integration quicker since you can substitute the standard result immediately.

After substitution, the integral becomes \( 5 \int \sinh(u) \, du \), which is straightforward as you apply the formula for the antiderivative. Completing the integration yields \( 5 \cosh(u) + C \), and replacing \( u \) back with its original value gives \( 5 \cosh \left( \frac{x}{5} \right) + C \).
Using these techniques ensures the integration process is manageable, effectively transforming complex functions into more solvable forms.