Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The integral is \( \frac{t^2}{2} \arctan t - \frac{1}{2}t + \frac{1}{2}\arctan t + C \).

Step 1: Identify Parts

In the integration by parts formula, we have \ \[ \int u \, dv = uv - \int v \, du \]. \First, we have to choose which part of the integral will be \( u \) and which part will be \( dv \). For this integral, let \( u = \arctan t \) and \( dv = t \, dt \).

Step 2: Differentiate u and Integrate dv

Differentiate \( u \) to find \( du \). \\[ du = \frac{1}{1+t^2} \, dt \] \Next, integrate \( dv \) to find \( v \). \\[ v = \frac{t^2}{2} \]

Step 3: Apply Integration by Parts Formula

Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula. \ \[ \int t \arctan t \, dt = \arctan t \frac{t^2}{2} - \int \frac{t^2}{2} \cdot \frac{1}{1+t^2} \, dt \]

Step 4: Simplify the Integral

Simplify the integral expression. \The current integral becomes: \ \[ \frac{t^2}{2} \arctan t - \frac{1}{2} \int \frac{t^2}{1+t^2} \, dt \]. \Notice that \( \frac{t^2}{1+t^2} = 1 - \frac{1}{1+t^2} \) so: \ \[ \frac{1}{2} \int \left( 1 - \frac{1}{1+t^2} \right) dt = \frac{1}{2} \left( \int 1 \, dt - \int \frac{1}{1+t^2} \, dt \right) \].

Step 5: Evaluate Remaining Integrals

Now solve the simplified integral: \ \[ \int 1 \, dt = t \] \\[ \int \frac{1}{1+t^2} \, dt = \arctan t \]. \Substituting back gives: \ \[ \frac{1}{2} \left(t - \arctan t \right) \].

Step 6: Final Solution

Combine all parts to obtain the final solution: \ \[ \frac{t^2}{2} \arctan t - \frac{1}{2} t + \frac{1}{2} \arctan t + C \] \where \( C \) is the constant of integration.

Key Concepts

definite integration

Definite integration is a fundamental tool for calculating the area under a curve between two specific limits. Unlike indefinite integrals that produce a family of functions with a constant of integration, definite integrals give a single value representing this area. They are expressed as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the lower and upper limits of integration, respectively.
To solve definite integrals, evaluate the antiderivative at these limits and subtract. For example, given an antiderivative \( F(x) \), the definite integral is \( F(b) - F(a) \).
In practice, the process involves first finding the indefinite integral, then computing these values at the specified bounds. This process preserves important properties like the signed area, where areas below the x-axis are considered negative. It's crucial to understand these principles to efficiently solve problems involving definite integrals.

arctan function

The arctan function, or inverse tangent function, is the inverse of the tangent function. It is denoted as \( \arctan(x) \) and gives the angle whose tangent is \( x \).
As with all inverse trigonometric functions, the arctan function has a restricted range, specifically from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\). This makes the function useful in many integration problems, especially when dealing with expressions involving \( 1/(1+x^2) \).
A key property of the arctan function is its derivative, which is \( \frac{d}{dx}(\arctan(x)) = \frac{1}{1+x^2} \). This derivative is often used to simplify integrals involving functions that look like the derivative of arctan. Understanding how to manipulate the arctan function and its properties is invaluable when confronting problems involving inverse trigonometric functions.

integration techniques

There are several techniques for integrating complex functions, each useful in different scenarios:

• Integration by Parts: Based on the product rule, it is ideal when dealing with products of functions. The formula is \( \int u \, dv = uv - \int v \, du \), where \( u \) and \( dv \) are parts of the original integral.
• Substitution: Used when an integral contains a composite function. By substituting \( u = g(x) \), this method simplifies the integrand to \( \int f(u) \, du \).
• Partial Fraction Decomposition: Utilized for rational functions, breaking them into simpler fractions which are easier to integrate.

Choosing the right technique depends on the type and complexity of the function you're working with. In our example, integration by parts was necessary because we had a product of a polynomial and an inverse trigonometric function. Mastery of these techniques ensures the ability to tackle a variety of integral problems efficiently.