/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 Compute the indefinite integral ... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 49

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

Expert verified
Question: Compute the indefinite integral of the vector-valued function \(\mathbf{r}(t) = \langle 2\cos t, 2\sin 3t, 4\cos 8t \rangle\). Answer: The indefinite integral of the given vector-valued function is \(\mathbf{R}(t) = \langle 2\sin t+c_1, -\frac{2}{3}\cos 3t+c_2, \frac{1}{2}\sin 8t+c_3 \rangle\), where \(c_1, c_2,\) and \(c_3\) are constants of integration.

Step by step solution

01

Compute the indefinite integral of each component

First, we'll find the indefinite integral of each component of the vector: 1. \(\int 2\cos t\, dt\) 2. \(\int 2\sin 3t\, dt\) 3. \(\int 4\cos 8t\, dt\) Remember to add a constant term to each indefinite integral.
02

Find the indefinite integral of the first component (2 cos t)

To integrate the first component: $$\int 2\cos t\, dt = 2\int \cos t\, dt$$ The integral of \(\cos t\) is \(\sin t\). So, the indefinite integral with respect to t is: $$2\int \cos t\, dt=2\sin t+c_1$$ where \(c_1\) is the constant of integration for this component.
03

Find the indefinite integral of the second component (2 sin 3t)

To integrate the second component: $$\int 2\sin 3t\, dt = 2\int \sin 3t\, dt$$ Using substitution, let \(u=3t\), so \(\frac{du}{dt} = 3\) or \(dt=\frac{1}{3}du\). Substituting into the integral, we get: $$2\int \sin u\frac{1}{3} du = \frac{2}{3}\int \sin u\, du$$ The integral of \(\sin u\) is \(-\cos u\). So, the indefinite integral with respect to u is: $$\frac{2}{3}\int \sin u\, du=-\frac{2}{3}\cos u+c_2$$ Substituting back \(u=3t\), we have: $$-\frac{2}{3}\cos 3t+c_2$$ where \(c_2\) is the constant of integration for this component.
04

Find the indefinite integral of the third component (4 cos 8t)

To integrate the third component: $$\int 4\cos 8t\, dt = 4\int \cos 8t\, dt$$ Using substitution, let \(v=8t\), so \(\frac{dv}{dt}=8\) or \(dt=\frac{1}{8}dv\). Substituting into the integral, we get: $$4\int \cos v\frac{1}{8} dv = \frac{1}{2}\int \cos v\, dv$$ The integral of \(\cos v\) is \(\sin v\). So, the indefinite integral with respect to v is: $$\frac{1}{2}\int \cos v\, dv=\frac{1}{2}\sin v+c_3$$ Substituting back \(v=8t\), we have: $$\frac{1}{2}\sin 8t+c_3$$ where \(c_3\) is the constant of integration for this component.
05

Combine the results

Now that we have the indefinite integral of each component, we can combine them back into a vector-valued function: $$\int \mathbf{r}(t)\, dt = \langle 2\sin t+c_1, -\frac{2}{3}\cos 3t+c_2, \frac{1}{2}\sin 8t+c_3 \rangle$$ So, the indefinite integral of the given vector-valued function is: $$\mathbf{R}(t) = \langle 2\sin t+c_1, -\frac{2}{3}\cos 3t+c_2, \frac{1}{2}\sin 8t+c_3 \rangle$$

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