Question

Find the integral by using the simplest method. Not all problems require integration by parts. $$ \int e^{x} \cos x d x $$

Short answer

\( \int e^x \cos x \, dx = \frac{1}{2} e^x (\sin x + \cos x) + C \)

Step-by-step solution

Identify the Integration Technique

Notice that this integral requires the use of integration by parts due to the multiplicative combination of an exponential and trigonometric function: \( e^x \) and \( \cos x \).

Set Up Integration by Parts Formula

Recall the integration by parts formula: \( \int u \, dv = uv - \int v \, du \). Choose \( u = e^x \) and \( dv = \cos x \, dx \). Thus, \( du = e^x \, dx \) and \( v = \int \cos x \, dx = \sin x \).

Apply the Integration by Parts Formula

Substitute the known elements into the formula: \( \int e^x \cos x \, dx = e^x \sin x - \int \sin x \, e^x \, dx \).

Repeat Integration by Parts for Remaining Integral

Now, apply integration by parts again on \( \int e^x \sin x \, dx \) by choosing \( u = e^x \) and \( dv = \sin x \, dx \). Then, \( du = e^x \, dx \) and \( v = -\cos x \).

Substitute Back into the Integral

Use integration by parts: \( \int e^x \sin x \, dx = -e^x \cos x - \int -\cos x \, e^x \, dx = -e^x \cos x + \int e^x \cos x \, dx \).

Solve the Equation

Notice you arrive back at the original integral. Let \( I = \int e^x \cos x \, dx \). From previous steps, we have \( I = e^x \sin x + e^x \cos x - I \). Simplifying yields \( 2I = e^x (\sin x + \cos x) \), thus \( I = \frac{1}{2} e^x (\sin x + \cos x) \).

Finalize the Solution

Don't forget the constant of integration! The final answer is \( \int e^x \cos x \, dx = \frac{1}{2} e^x (\sin x + \cos x) + C \), where \( C \) is the constant of integration.

Key concepts

Exponential Functions

Exponential functions are a fundamental part of calculus and mathematics in general. They are of the form \( e^x \), where \( e \) is a constant approximately equal to 2.71828. This constant is unique because it is the base of the natural logarithm.
Unlike linear and polynomial functions, exponential functions grow at an increasingly rapid rate as \( x \) increases. For example:

• \( x = 1 \), then \( e^x = e \)
• \( x = 2 \), then \( e^x = e^2 \)
• \( x = -1 \), then \( e^x = \frac{1}{e} \)

In calculus, exponential functions are important due to their unique property of having a derivative that is proportional to the function itself.
Specifically, the derivative of \( e^x \) is still \( e^x \), making these functions relatively simple to handle in differential equations and integrations. This characteristic is a key aspect in the choice of functions when solving problems with techniques like integration by parts.

Trigonometric Functions

Trigonometric functions, which include \( \sin x \) and \( \cos x \), are the building blocks of trigonometry. These functions relate angles of a triangle to the lengths of its sides.
But importantly, they are defining elements in calculus, especially when dealing with periodic phenomena.

• \( \sin x \) represents the y-coordinate of a point on the unit circle associated with an angle \( x \).
• \( \cos x \) represents the x-coordinate of the same point.

Integrating trigonometric functions often requires special techniques, as their derivatives cycle through each other: the derivative of \( \sin x \) is \( \cos x \), and the derivative of \( \cos x \) is \(-\sin x \).
In problems involving exponential and trigonometric functions, such as \( e^x \cos x \), the integration by parts method helps manage these cyclic behaviors by carefully selecting which part of the integrand to differentiate and which part to integrate.

Definite and Indefinite Integrals

Integrals in calculus come in two main types: definite and indefinite.
Indefinite integrals, like those used in the problem, represent a family of functions and are denoted by \( \int f(x) \, dx \). The process is about finding a general form of antiderivative, including a constant \( C \), since any constant added to a function goes away after differentiation. For example:

• \( \int e^x \, dx = e^x + C \)
• \( \int \cos x \, dx = \sin x + C \)

Definite integrals, on the other hand, evaluate the area under a curve from one point to another, expressed as \( \int_{a}^{b} f(x) \, dx \).
In this exercise, we are concerned with an indefinite integral, manipulating the equation until it reflects a neat expression with an added constant. Integration by parts allows us to transform the problem into more manageable parts, solving one piece at a time.