Simplify and enter your final answer. * T/6 (sec t tan t)i + (tan t)j + (2 sin t cos t)k dt = Your answer cannot be understood or graded. More Information
Simplify and enter your final answer. * T/6 (sec t tan t)i + (tan t)j + (2 sin t cos t)k dt = Your answer cannot be understood or graded. More Information
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![## Find the Indefinite Integral
### Problem:
Find the indefinite integral:
\[
\int_{0}^{\pi/6} \left( (\sec t \tan t) \mathbf{i} + (\tan t) \mathbf{j} + (2 \sin t \cos t) \mathbf{k} \right) \, dt
\]
### Part 1 of 3:
Integrate the given integral with respect to \( t \) on a component-by-component basis.
- \[
\int_{0}^{\pi/6} \left( (\sec t \tan t) \mathbf{i} + (\tan t) \mathbf{j} + (2 \sin t \cos t) \mathbf{k} \right) \, dt
\]
Breaking down the components:
1. \[
\int_{0}^{\pi/6} (\sec t \tan t) \, dt \, \mathbf{i} = \left[\sec(t)\right]_{0}^{\pi/6} \mathbf{i}
\]
2. \[
\int_{0}^{\pi/6} (\tan t) \, dt \, \mathbf{j} = \left[ \ln|\sec(t)|\right]_{0}^{\pi/6} \mathbf{j}
\]
3. \[
\int_{0}^{\pi/6} (2 \sin t \cos t) \, dt \, \mathbf{k} = \left[ \sin^2(t)\right]_{0}^{\pi/6} \mathbf{k}
\]
**Components:**
- \(\sec(t)\) evaluated from 0 to \(\pi/6\)
- \(\ln|\sec(t)|\) evaluated from 0 to \(\pi/6\)
- \(\sin^2(t)\) evaluated from 0 to \(\pi/6\)
### Part 2 of 3:
Evaluate the expressions.
- \[
\left( \sec \frac{\pi}{6} - \sec 0 \right)\mathbf{i} + \left( \ln \sec \frac{\pi}{6} - \ln \sec 0 \right)\mathbf{j} + (\sin^2 \frac{\pi}{6} - \sin^2 0)\mathbf{k}
\]
Calculations:
- \(\frac{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc08bab33-dfc3-4475-87c9-eef4dfb69a19%2F57763d57-15cf-4f22-bd32-79fa98bd96e5%2Fyacg0at_processed.png&w=3840&q=75)
Transcribed Image Text:## Find the Indefinite Integral
### Problem:
Find the indefinite integral:
\[
\int_{0}^{\pi/6} \left( (\sec t \tan t) \mathbf{i} + (\tan t) \mathbf{j} + (2 \sin t \cos t) \mathbf{k} \right) \, dt
\]
### Part 1 of 3:
Integrate the given integral with respect to \( t \) on a component-by-component basis.
- \[
\int_{0}^{\pi/6} \left( (\sec t \tan t) \mathbf{i} + (\tan t) \mathbf{j} + (2 \sin t \cos t) \mathbf{k} \right) \, dt
\]
Breaking down the components:
1. \[
\int_{0}^{\pi/6} (\sec t \tan t) \, dt \, \mathbf{i} = \left[\sec(t)\right]_{0}^{\pi/6} \mathbf{i}
\]
2. \[
\int_{0}^{\pi/6} (\tan t) \, dt \, \mathbf{j} = \left[ \ln|\sec(t)|\right]_{0}^{\pi/6} \mathbf{j}
\]
3. \[
\int_{0}^{\pi/6} (2 \sin t \cos t) \, dt \, \mathbf{k} = \left[ \sin^2(t)\right]_{0}^{\pi/6} \mathbf{k}
\]
**Components:**
- \(\sec(t)\) evaluated from 0 to \(\pi/6\)
- \(\ln|\sec(t)|\) evaluated from 0 to \(\pi/6\)
- \(\sin^2(t)\) evaluated from 0 to \(\pi/6\)
### Part 2 of 3:
Evaluate the expressions.
- \[
\left( \sec \frac{\pi}{6} - \sec 0 \right)\mathbf{i} + \left( \ln \sec \frac{\pi}{6} - \ln \sec 0 \right)\mathbf{j} + (\sin^2 \frac{\pi}{6} - \sin^2 0)\mathbf{k}
\]
Calculations:
- \(\frac{
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