Question

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

Short answer

The integral is \( x \cosh x - \sinh x + C \).

Step-by-step solution

Identify Parts for Integration by Parts Formula

The formula for integration by parts is: \[ \int u \, dv = uv - \int v \, du \] We need to identify which part of the integral will be \( u \) and which part will be \( dv \). Let \( u = x \) and \( dv = \sinh x \, dx \).

Differentiate and Integrate

Differentiate \( u \) and integrate \( dv \): \[ du = dx \] \[ v = \int \sinh x \, dx = \cosh x \] Now we have all the components needed for the integration by parts formula.

Apply Integration by Parts Formula

Substitute \( u \), \( du \), \( v \), and \( dv \) into the formula: \[ \int x \, \sinh x \, dx = x \cosh x - \int \cosh x \, dx \] Simplify the expression.

Solve Remaining Integral

Evaluate \( \int \cosh x \, dx \): \[ \int \cosh x \, dx = \sinh x \] Substitute back into the equation: \[ x \cosh x - \sinh x + C \] where \( C \) is the constant of integration.

Key concepts

Integration by Parts

Integration by parts is a technique used in calculus to integrate products of functions. Based on the product rule for differentiation, it's useful when dealing with two functions multiplied together. The formula for integration by parts is: \[ \int u \, dv = uv - \int v \, du \] To use integration by parts, follow these steps:

• Select which part of the integrand will be \( u \) and which will be \( dv \). A good rule of thumb is to choose \( u \) to be a function that becomes simpler when differentiated.
• Differentiate \( u \) to find \( du \) and integrate \( dv \) to find \( v \).
• Plug these into the integration by parts formula.
• Perform the remaining integration if necessary.

In our original exercise, we chose \( u = x \) because differentiating \( x \) simplifies it to \( 1 \), while \( dv = \sinh x \, dx \) was selected since its integral, \( \cosh x \), is straightforward. After simplifying, you'll often find that the integration becomes more manageable.

Hyperbolic Functions

Hyperbolic functions are analogs of the ordinary trigonometric functions but are based on hyperbolas instead of circles. They find applications in many areas of mathematics, including calculus and complex analysis. Key hyperbolic functions include \( \sinh x \) (hyperbolic sine) and \( \cosh x \) (hyperbolic cosine).The definitions are as follows:

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

These functions have properties similar to the trigonometric ones:

• The derivative of \( \sinh x \) is \( \cosh x \).
• The derivative of \( \cosh x \) is \( \sinh x \).
• A key identity: \( \cosh^2 x - \sinh^2 x = 1 \).

Understanding these functions lays the groundwork for solving integrals involving hyperbolic functions, as their derivatives integrate nicely just like in the example from the exercise.

Definite and Indefinite Integrals

Integrals in calculus are classified into two main types: definite and indefinite integrals. Understanding these is crucial for solving a variety of problems.Indefinite integrals, such as \( \int f(x) \, dx \), represent a family of functions and include an arbitrary constant \( C \) because there is no specified interval of integration. They answer the question: "What function has a derivative that is \( f(x) \)?"Definite integrals, written as \( \int_a^b f(x) \, dx \), compute the net area under the curve \( f(x) \) from \( x=a \) to \( x=b \). This evaluation gives a specific numerical value, not a new function.For both types:

• Antiderivatives are the basis. Finding an antiderivative is the key step for indefinite integration.
• The Fundamental Theorem of Calculus links these concepts by stating that if \( F(x) \) is an antiderivative of \( f(x) \), then \( \int_a^b f(x) \, dx = F(b) - F(a) \).

In our example, we computed an indefinite integral involving hyperbolic functions, yielding \( x \cosh x - \sinh x + C \) as the result, where \( C \) is the constant of integration indicative of its indefinite nature.