Question

$$ \text { use integration by parts to evaluate each integral. } $$ $$ \int x \cosh x d x $$

Short answer

\( \int x \cosh x \, dx = x \sinh x - \cosh x + C \)

Step-by-step solution

Identify Functions

Identify the parts to use for integration by parts, applying the formula: \( \int u \, dv = uv - \int v \, du \). Choose \( u = x \) so that \( du = dx \). Let \( dv = \cosh x \, dx \), so that we find \( v \) by integration.

Differentiate and Integrate

Differentiate \( u = x \) to get \( du = dx \). Integrate \( dv = \cosh x \, dx \) to find \( v = \sinh x \), since the derivative of \( \sinh x \) is \( \cosh x \).

Apply Integration by Parts Formula

Substitute into the integration by parts formula: \( \int x \cosh x \, dx = x \sinh x - \int \sinh x \, dx \).

Integrate Remaining Integral

Evaluate the simpler integral \( \int \sinh x \, dx \). The integral of \( \sinh x \) is \( \cosh x \), so \( \int \sinh x \, dx = \cosh x + C \) (where \( C \) is the constant of integration).

Substitute Back and Simplify

Substitute \( \cosh x \) from the previous step back into the equation: \( \int x \cosh x \, dx = x \sinh x - \cosh x + C \).

Key concepts

Hyperbolic Functions

Hyperbolic functions, such as the hyperbolic cosine and hyperbolic sine, play a crucial role in calculus, similar to their trigonometric counterparts. They are defined based on exponential functions and have unique properties that make them useful in various contexts, especially when dealing with differential equations and integration problems.

- **Hyperbolic Cosine (\( \cosh x \))**: This function is defined as \( \cosh x = \frac{e^x + e^{-x}}{2} \). It describes the shape of a catenary, which is the curve formed by a hanging flexible chain or cable under gravity.
- **Hyperbolic Sine (\( \sinh x \))**: Its definition is \( \sinh x = \frac{e^x - e^{-x}}{2} \). When you differentiate \( \cosh x \), you get \( \sinh x \), and the reverse is true for the derivative of \( \sinh x \), which is \( \cosh x \).

These functions satisfy identities similar to trigonometric functions, but with subtle differences that are crucial to remember when solving calculus problems, like integration by parts.

Integral Calculus

Integral calculus focuses on the concept of the integral, which is like a continuous sum. It's used to find areas under curves, volumes, central points, and many more quantities. The fundamental operations of calculus involve taking derivatives and integrals.

- **Definite Integrals**: They calculate the accumulated quantity within a given interval and are represented as \( \int_{a}^{b} f(x) \, dx \).
- **Indefinite Integrals**: These provide general forms of antiderivatives and are written as \( \int f(x) \, dx = F(x) + C \), where \( F(x) \) is the antiderivative and \( C \) is the constant of integration.

Integration by parts is a special technique in integral calculus. This method is particularly helpful when the integral can be decomposed into a product of functions where one part simplifies upon differentiation, and the other can easily be integrated.

Calculus Problem Solving

Solving calculus problems often involves recognizing patterns and applying specific mathematical techniques like integration by parts. This technique follows the principle of the product rule for differentiation but applies it in reverse to simplify complex integrals.

- **Integration by Parts Formula**: The formula is \( \int u \, dv = uv - \int v \, du \). It's most effective when chosen parts for \( u \) and \( dv \) allow an easy computation of \( du \) and \( v \).
- **Selection of \( u \) and \( dv \)**: Often, \( u \) is a function that becomes simpler upon differentiation, like polynomials, and \( dv \) is chosen as a function whose integral is known.

In our example, integrating \( \int x \cosh x \, dx \), starting with \( u = x \) and \( dv = \cosh x \, dx \) effectively simplifies the problem. Solving calculus challenges like this one enhances logical thinking and mathematical understanding, ultimately providing clear paths through complex integrals.