Problem 38 - Calculus Early Transcendentals

Question

Evaluate the integral. $$ \int(\cos \pi t \mathbf{i}+\sin \pi t \mathbf{j}+t \mathbf{k}) d t $$

Step-by-Step Solution

Answer

\( \frac{1}{\pi} \sin(\pi t) \mathbf{i} - \frac{1}{\pi} \cos(\pi t) \mathbf{j} + \frac{t^2}{2} \mathbf{k} + \mathbf{C} \)

Step 1: Break Down the Vector Integral

We start by recognizing that this is a vector integral with components inside. Thus, we can split it into three separate integrals, one for each component:\[ \int ig( (\cos \pi t) \mathbf{i} + (\sin \pi t) \mathbf{j} + t \mathbf{k} \big) \, dt = \int (\cos \pi t) \, dt \, \mathbf{i} + \int (\sin \pi t) \, dt \, \mathbf{j} + \int t \,dt \, \mathbf{k}. \]

Step 2: Evaluate the Integral for the i-component

We need to find \( \int (\cos \pi t) \, dt \). This requires a u-substitution where \( u = \pi t \) and \( du = \pi \, dt \), so \( dt = \frac{du}{\pi} \). The integral becomes:\[ \int \cos \pi t \, dt = \frac{1}{\pi} \int \cos u \, du = \frac{1}{\pi} \sin u = \frac{1}{\pi} \sin(\pi t). \]Thus, the i-component of the vector will be \( \frac{1}{\pi} \sin(\pi t) \mathbf{i} \).

Step 3: Evaluate the Integral for the j-component

Now for \( \int (\sin \pi t) \, dt \). Again, use substitution \( u = \pi t \), and \( dt = \frac{du}{\pi} \), transforming the integral:\[ \int \sin \pi t \, dt = \frac{1}{\pi} \int \sin u \, du = -\frac{1}{\pi} \cos u = -\frac{1}{\pi} \cos(\pi t). \]So the j-component of the vector is \(-\frac{1}{\pi} \cos(\pi t) \mathbf{j} \).

Step 4: Evaluate the Integral for the k-component

Finally, find \( \int t \, dt \), which is straightforward:\[ \int t \, dt = \frac{t^2}{2}. \]This gives us the k-component of the vector: \( \frac{t^2}{2} \mathbf{k} \).

Step 5: Combine the Components

Now, we need to combine all these integral results to form the solution for the entire vector integral:\[ \int (\cos \pi t \mathbf{i} + \sin \pi t \mathbf{j} + t \mathbf{k}) \, dt = \frac{1}{\pi} \sin(\pi t) \mathbf{i} - \frac{1}{\pi} \cos(\pi t) \mathbf{j} + \frac{t^2}{2} \mathbf{k} + \mathbf{C}, \]where \( \mathbf{C} \) is the constant of integration, which represents an arbitrary constant vector.

Key Concepts

Vector Integral

In vector calculus, a vector integral involves integrating a vector function. This can seem a bit complex at first because it deals with multiple components. A vector function can be broken down into its scalar component parts. Each of these parts—often three in 3D space, represented as \( \mathbf{i}, \mathbf{j}, \mathbf{k} \)—can be integrated separately. For example, consider the integral \( \int ( \cos \pi t \mathbf{i} + \sin \pi t \mathbf{j} + t \mathbf{k} ) \ dt \). You can split this into three separate integrals:

• \( \int (\cos \pi t) \ dt \mathbf{i} \)
• \( \int (\sin \pi t) \ dt \mathbf{j} \)
• \( \int t \ dt \mathbf{k} \)

Each component integral can then be solved individually, and their results recombined to form the solution for the whole vector integral. This technique allows for simplified computation and understanding of the function's behavior.

U-substitution Integration

Sometimes, integrals involve functions composed with another function, called a composition of functions. U-substitution is a technique used to simplify the integration of these compositions. The basic idea is to replace part of the integral with a variable \( u \), which stands for a function inside the integral. Consider the integral \( \int (\cos \pi t) \ dt \). We set \( u = \pi t \), leading to \( du = \pi \ dt \) or \( dt = \frac{du}{\pi} \). This substitution transforms the integral into an easier form:

• \( \int \cos \pi t \ dt \rightarrow \frac{1}{\pi} \int \cos u \ du \)

Substituting back the original variables after integration gives the result in terms of \( t \). It’s crucial to switch back to original variables to keep the solution relevant to the problem context. This technique often simplifies the integration of trigonometric functions and other compositions.

Integration of Trigonometric Functions

Trigonometric functions are central to many integrals in calculus. Understanding their basic integrals can be very useful. Two fundamental integrals are:

• \( \int \cos x \ dx = \sin x + C \)
• \( \int \sin x \ dx = -\cos x + C \)

Applying these rules alongside other calculus methods like u-substitution can help solve more complex integrals. For example, solving \( \int (\sin \pi t) \ dt \) starts by letting \( u = \pi t \), which simplifies the integral to:

• \( \int \sin u \ du \to -\cos u \to -\frac{1}{\pi} \cos(\pi t) \)

Remember to always incorporate a constant of integration, \( C \), in indefinite integrals. This constant represents any potential vertical shift of the function along the y-axis, vital in making sure the integral reflects all possible solutions to the problem.