Question

use integration by parts to derive the given formula. $$ \int e^{\alpha z} \cos \beta z d z=\frac{e^{\alpha z}(\alpha \cos \beta z+\beta \sin \beta z)}{\alpha^{2}+\beta^{2}}+C $$

Short answer

Integrate by parts twice to verify \( \int e^{\alpha z} \cos \beta z \, dz = \frac{e^{\alpha z}(\alpha \cos \beta z + \beta \sin \beta z)}{\alpha^2 + \beta^2} + C \).

Step-by-step solution

Identify Functions

In this integration problem, use integration by parts, where \( u = e^{\alpha z} \) and \( dv = \cos \beta z \; dz \). This means that we need to find \( du \) and \( v \).

Differentiate and Integrate Functions

Calculate the differential \( du \):\( du = \alpha e^{\alpha z} dz \). Integrate \( dv \) to find \( v \): \( v = \frac{1}{\beta} \sin \beta z \).

Apply Integration by Parts Formula

Use the integration by parts formula, \( \int u \, dv = uv - \int v \, du \). Substitute the identified and calculated values to get \( \int e^{\alpha z} \cos \beta z \, dz = e^{\alpha z} \cdot \frac{1}{\beta} \sin \beta z - \int \frac{1}{\beta} \sin \beta z \cdot \alpha e^{\alpha z} \cdot dz \).

Simplify the Equation

Simplify the first term: \( \frac{1}{\beta} e^{\alpha z} \sin \beta z \). For the second integral: \( \int \frac{\alpha}{\beta} e^{\alpha z} \sin \beta z \, dz \).

Perform Another Integration by Parts

Apply integration by parts again to \( \int e^{\alpha z} \sin \beta z \, dz \), choosing \( u = e^{\alpha z} \) and \( dv = \sin \beta z \, dz \). Differentiate: \( du = \alpha e^{\alpha z} dz \). Integrate: \( v = -\frac{1}{\beta} \cos \beta z \).

Substitute and Simplify

Substitute into integration by parts formula: \( \int e^{\alpha z} \sin \beta z \, dz = -\frac{1}{\beta} e^{\alpha z} \cos \beta z + \int \frac{\alpha}{\beta} e^{\alpha z} \cos \beta z \, dz \).

Derive the Final Formula

The original integral becomes\[ \frac{1}{\beta} e^{\alpha z} \sin \beta z - \frac{1}{\beta}(-\frac{1}{\beta} e^{\alpha z} \cos \beta z + \int \frac{\alpha}{\beta} e^{\alpha z} \cos \beta z dz) \]. Simplify to verify given solution: \( \frac{e^{\alpha z}(\alpha \cos \beta z + \beta \sin \beta z)}{\alpha^2 + \beta^2} + C \), matching the given formula.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve a function and its derivatives. They play a crucial role in describing various real-life phenomena, such as the rate of change of populations, the spread of diseases, or the dynamics of ecosystems. In our exercise, although it's mainly focused on integral calculus, differential equations provide the underlying principle for transformations often used in calculus.
They are typically classified into ordinary differential equations (ODEs), involving derivatives with respect to a single variable, and partial differential equations (PDEs), involving derivatives with respect to multiple variables. The formula derived in the original exercise, using integration by parts, reveals the relationship between an exponential function and a trigonometric function. This can be considered a form of solution to a differential equation.
Understanding these relationships is crucial as they can explain the behavior of more complex systems. By integrating a function, you're essentially "undoing" the differentiation to find the original function, which is essential for solving differential equations.

Transcendental Functions

Transcendental functions are a class of functions that go beyond algebraic functions. They cannot be expressed with a finite sequence of the operations of addition, multiplication, and root extraction. Common examples of transcendental functions are exponential functions, logarithmic functions, and trigonometric functions.
In our problem, we're dealing with the exponential function, represented here as \( e^{\alpha z} \), and trigonometric functions like \( \cos \beta z \) and \( \sin \beta z \). These functions are central to many areas of mathematics and physics because they often describe non-linear growth, oscillations, and waves.
Importantly, transcendental functions often require specialized techniques for integration, such as integration by parts or substitution, as seen in this exercise. This is because they do not adhere to the customary algebraic rules, thus necessitating a deep understanding of their behavior and properties for integration and solving equations involving them.

Integral Calculus

Integral calculus is a branch of calculus concerned with the theory and applications of integrals. It is used to calculate quantities like areas, volumes, central points, and many more. Integration by parts is a fundamental technique in integral calculus, aiding in the evaluation of more complex integrals.
The integration by parts formula, \( \int u \, dv = uv - \int v \, du \), is a powerful tool derived from the product rule for differentiation. In our exercise, this formula is used twice: first to tackle the initial integral of an exponential function and a cosine function, then to address the remaining integral of the sine function.
By breaking down complex integrals into more manageable parts, integration by parts helps in systematically solving integrals that involve products of functions. Understanding and applying this technique is essential in physics, engineering, and other fields where calculating the area under curves is frequent.