Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-36, use integration by parts to evaluate each integral. $$ \int \arctan x d x $$

Step-by-Step Solution

Answer

\( \int \arctan x \, dx = x \arctan x - \frac{1}{2} \ln(1+x^2) + C \).

Step 1: Identify parts for Integration by Parts

In integration by parts, we need to identify two parts: a function to differentiate (let's call it \( u \)) and a function to integrate (let's call it \( dv \)). Here, let's choose \( u = \arctan x \) and \( dv = dx \).

Step 2: Differentiate and Integrate Parts

Differentiate \( u \) and integrate \( dv \). We get \( du = \frac{1}{1+x^2} dx \) and \( v = x \).

Step 3: Apply the Integration by Parts Formula

The formula for integration by parts is: \[ \int u \, dv = uv - \int v \, du \]Substitute the values: \[ \int \arctan{x} \, dx = x \arctan{x} - \int x \cdot \frac{1}{1+x^2} \, dx \]

Step 4: Simplify and Integrate the Remaining Integral

Simplify the integral: \[ \int \frac{x}{1+x^2} \, dx = \int \frac{1}{2} \cdot \frac{2x}{1+x^2} \, dx \]Recognize \( \frac{d}{dx}(1+x^2) = 2x \) so let \( w = 1+x^2 \) which simplifies to \( dw = 2x \, dx \). Thus the integral becomes \[ \frac{1}{2} \int \frac{1}{w} \, dw \].

Step 5: Solve the Integral

This integral \( \int \frac{1}{w} \, dw \) results in \( \ln|w| \), so the expression becomes: \[ \frac{1}{2} \ln|1+x^2| \].

Step 6: Compile the Final Solution

Combine the results from integration by parts and the simplified integral:\[ x \arctan{x} - \frac{1}{2} \ln(1+x^2) + C \]Here, \( C \) is the constant of integration.

Key Concepts

Integral Calculus

Integral calculus is a core part of mathematics that focuses on the process of integration. It involves finding the integral of functions, which essentially means finding the area under curves or the accumulation of quantities. In the context of this exercise, we're exploring how to integrate functions involving specific expressions like the arctan function. Integration is often the reverse process of differentiation, and various methods, like substitution and integration by parts, help solve different integrals.

For this specific exercise, we utilize integration by parts, a special technique used when integrating the product of two functions. This method is best applied when you can identify parts of the integral that can be simplified through differentiation and direct integration, respectively. The formula for integration by parts is given by:

• \( \int u \, dv = uv - \int v \, du \)

Understanding when and how to apply these techniques is crucial for tackling various complex integrals in integral calculus.

Arctan Function

The arctan function, short for arctangent, is an inverse trigonometric function. It represents the angle whose tangent is a given number. In mathematical terms, if \( y = \arctan(x) \), then \( \tan(y) = x \). The arctan function, like other inverse trigonometric functions, has a specific derivative which is useful in calculus.

• Derivative: \( \frac{d}{dx} \arctan(x) = \frac{1}{1+x^2} \).

In this exercise, we chose \( u = \arctan(x) \) in the integration by parts formula. This choice leveraged the straightforward differentiation of the arctan function. By differentiating the arctan function, we obtain a simpler expression involving \( du \).

The arctan function is important because it frequently appears in integration problems. Its unique properties and well-defined derivative make it a common component when studying integral calculus.

Differentiation

Differentiation is the calculus process of finding the derivative of a function. The derivative represents the rate of change of a function with respect to its variable. When using integration by parts, differentiation plays a crucial role.

For integration by parts, you must differentiate the chosen \( u \) part of the function. In our exercise, we differentiated the arctan function:

• \( du = \frac{1}{1+x^2} \) dx

This differentiation leads us to the related integral that will be subtracted in the integration by parts formula. Differentiation provides the necessary change in one of the components, helping simplify the entire integration problem.

It's important to understand how differentiation interacts with integration because they are inverse operations. Mastering both gives you tools for analyzing and solving complex mathematical problems.

Definite Integrals

While definite integrals were not specifically part of this exercise, understanding them is beneficial. A definite integral computes the accumulated quantity or area between two bounds. Unlike indefinite integrals, which have a +C constant, definite integrals provide a specific number.

When given a definite integral, such as \( \int_{a}^{b} \arctan{x} \ dx \), the process involves evaluating the antiderivative—found using methods like integration by parts—at both limits. The area or accumulated value is then the difference of these evaluations:

• \( F(b) - F(a) \)

Understanding definite integrals helps in contexts where specific numeric results are required and also illuminates areas or total quantities, such as probabilities in statistics or masses in physics.

By mastering both definite and indefinite integrals, you enhance your ability to solve a broad array of mathematical challenges.