Question

Compute the indefinite integral of the following functions. $$\mathbf{r}(t)=\langle 2 \cos t, 2 \sin 3 t, 4 \cos 8 t\rangle$$

Short answer

Question: Find the indefinite integral of the vector-valued function $$\mathbf{r}(t)=\langle 2 \cos t, 2 \sin 3 t, 4 \cos 8 t\rangle$$. Answer: $$\mathbf{R}(t)=\left\langle 2 \sin t + C_1, -\frac{2}{3}\cos 3 t + C_2, \frac{1}{2} \sin 8 t + C_3 \right\rangle$$ where \(C_1\), \(C_2\), and \(C_3\) are integration constants.

Step-by-step solution

Analyze the given vector function

The given vector function is: $$\mathbf{r}(t)=\langle 2 \cos t, 2 \sin 3 t, 4 \cos 8 t\rangle$$ We want to find the indefinite integral of that function, which means that we need to integrate each component separately.

Integrate the first component

We need to integrate: $$\int 2 \cos t \, dt$$ The integral of 2cos(t) with respect to t is 2sin(t), so we find: $$\int 2 \cos t \, dt = 2 \sin t + C_1$$ Where \(C_1\) is the integration constant for the first component.

Integrate the second component

Now we need to integrate: $$\int 2 \sin 3 t \, dt$$ Using the chain rule, we get: $$\int 2 \sin 3 t \, dt = -\frac{2}{3}\cos 3 t + C_2$$ Where \(C_2\) is the integration constant for the second component.

Integrate the third component

Finally, we integrate: $$\int 4 \cos 8 t \, dt$$ Using the chain rule again, we get: $$\int 4 \cos 8 t \, dt = \frac{1}{2} \sin 8 t + C_3$$ Where \(C_3\) is the integration constant for the third component.

Combine the results

Now that we have integrated each component, we combine them to find the antiderivative vector: $$\mathbf{R}(t)=\left\langle 2 \sin t + C_1, -\frac{2}{3}\cos 3 t + C_2, \frac{1}{2} \sin 8 t + C_3 \right\rangle$$

Key concepts

Vector Integration

Vector integration involves finding the antiderivative of a vector-valued function by integrating each of its components separately. Imagine a vector as a collection of individual functions packaged together. For instance, consider the vector function \( \mathbf{r}(t) = \langle 2 \cos t, 2 \sin 3t, 4 \cos 8t \rangle \). This vector is essentially made of three individual functions: \( 2\cos t \), \( 2\sin 3t \), and \( 4\cos 8t \). To determine the indefinite integral of the vector, you separately integrate each function.

• Step 1: Identify each component of the vector.
• Step 2: Integrate each component individually.
• Step 3: Gather the results to form the antiderivative vector.

This method simplifies complex problems, allowing you to focus on integrating simpler, individual functions.

Trigonometric Functions

Trigonometric functions such as sine and cosine are essential in calculus, and understanding their integrals is crucial. These functions periodically oscillate, and their integrals help calculate areas, among other applications. Let's look at some basic integrals of trig functions:

• The integral of \( \cos(t) \) is \( \sin(t) \).
• The integral of \( \sin(t) \) is \( -\cos(t) \).

In the example exercise, these properties make it easier to handle the function's components, like \( 2\cos t \) and \( 2\sin 3t \). Recognizing these basic patterns speeds up problem solving and aids comprehension of more elaborate calculus challenges.

Chain Rule

The chain rule is a method in calculus used primarily for differentiating the composition of functions. However, its principles can be reversed and applied to integration, commonly known as "u-substitution." When integrating functions involving compositions like \( 2\sin 3t \) and \( 4\cos 8t \), you use integration methods derived from the chain rule.The chain rule in differentiation states:

• If \( y = f(g(x)) \)
• Then \( dy/dx = f'(g(x)) \cdot g'(x) \)

In integration, you do the reverse by identifying a part of the function (often the inside function) and considering its derivative, simplifying the process of finding the integral. In our example, notice how the factor \( \frac{1}{3} \) comes up in \( \int 2\sin 3 t \, dt = -\frac{2}{3}\cos 3t + C_2 \), adjusting for the inner function derivative \( 3 \).

Integration Constant

The integration constant, often denoted as \( C \), is a vital part of indefinite integrals. When you integrate a function, you get a family of functions because of this constant. Each value of \( C \) provides a specific solution differing by a vertical shift on the function's graph.For instance, when integrating \( 2\cos t \), the result is \( 2\sin t + C_1 \). The constant \( C_1 \) accounts for any constant vertical displacement in the potential solutions.

• The constant accounts for the indefinite nature of integration, distinguishing from initial values.
• This is why when integrating vector functions, you have a separate constant for each component, like \( C_1, C_2, \) and \( C_3 \).

Understanding this constant's role helps ensure complete solutions in calculus problems, providing you with the complete set of possible antiderivatives.