Problem 7 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int x \cos x d x$$

Step-by-Step Solution

Answer

Question: Evaluate the indefinite integral: $$\int x \cos x dx$$ Answer: The indefinite integral is: $$\int x \cos x dx = x \sin x + \cos x + C$$

Step 1: Apply integration by parts formula

Recall the integration by parts formula: $$\int u d v = u v - \int v d u$$ First, we need to choose our functions "u" and "dv" from the given integral. Usually, we pick "u" to be the function that becomes simpler when differentiated, and "dv" to be the function that doesn't get more complicated when integrated. In this case, let's choose: $$u = x$$ $$d v = \cos x dx$$

Step 2: Differentiate u and integrate dv

Next, we need to differentiate "u" and integrate "dv": $$d u = \frac{d(x)}{dx} dx = dx$$ $$v = \int \cos x dx = \sin x$$

Step 3: Apply the formula

Now we plug our functions "u", "v", "du", and "dv" back into the integration by parts formula: $$\int x \cos x dx = \int u dv = uv - \int v du = x \sin x - \int \sin x dx$$

Step 4: Evaluate the remaining integral

Now we need to integrate the remaining integral $$\int \sin x dx$$, which is: $$\int \sin x dx = -\cos x + C$$ where "C" is the integration constant.

Step 5: Combine the results

Finally, substitute the result of the remaining integral back into the expression we obtained in Step 3: $$\int x \cos x dx = x \sin x - (-\cos x + C)$$ Thus, the result of the given integral is: $$\int x \cos x dx = x \sin x + \cos x + C$$

Key Concepts

Integrals

Integrals are a fundamental concept in calculus. They can be thought of as a way to accumulate quantities. This is somewhat like adding up many small pieces to get a total. The process of integration combines these small amounts under a curve to compute the total area beneath it.

• Understanding integrals involves comprehending how they reverse the process of differentiation.
• There are different methods to evaluate integrals, such as substitution, partial fractions, and integration by parts.
• The symbol for integration is \( \int \), and it is often accompanied by limits of integration for evaluating definite integrals.

Integration by parts is a specific technique that uses the integration by parts formula \( \int u \: dv = uv - \int v \: du \) to integrate products of functions.
The choice of "u" and "dv" is crucial for simplifying the integral. In our example, setting \( u = x \) and \( dv = \cos x \, dx \) simplifies the integration process.

Differential Calculus

Differential calculus focuses on the concept of the derivative, which represents the rate of change or the slope of a curve. Differentiation is the process of finding the derivative of a function.

• Useful for finding tangents to curves or optimizing functions (finding maxima or minima).
• Derivatives are denoted by \( f'(x) \) or \( \frac{dy}{dx} \).

Differentiation aids integration, particularly integration by parts. In our example, we first differentiate \( u = x \) to find \( du = dx \).
This step is critical as it allows us to rearrange the components of the integral in integration by parts.The purpose is to break down complex integrals into simpler pieces that are straightforward to solve.

Definite and Indefinite Integrals

Integration can yield two types of results: definite and indefinite integrals. Each serves its purpose in calculus.

• Indefinite Integrals: Represent the general form of antiderivatives of a function. They include a constant of integration \( C \), because differentiating a constant yields zero. For instance, \( \int \sin x \, dx = -\cos x + C \).
• Definite Integrals: Calculate the area under a curve between two bounds. They result in a numerical value. Notably, definite integrals do not require an integration constant. For example, \( \int_{a}^{b} f(x) \, dx \) evaluates to the precise area from \( x = a \) to \( x = b \).

In our step-by-step solution, we focused on finding the indefinite integral of \( x \cos x \), denoting the general antiderivative. The inclusion of "+ C" indicates it can encompass a variety of functions that differentiate to the same form.