3t – 9 Integrate using the trigonometric substitution method dt J? - 6t +13

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...
icon
Related questions
Question
### Problem Statement
**Integrate using the trigonometric substitution method:**

\[ \int \frac{3t - 9}{t^2 - 6t + 13} \, dt \]

### Explanation

In this problem, our goal is to evaluate the integral of the function \(\frac{3t - 9}{t^2 - 6t + 13}\) with respect to \(t\) by employing the trigonometric substitution method.

This technique is often used for integrals involving algebraic expressions that can be transformed into trigonometric forms, making them easier to integrate. The expression \(t^2 - 6t + 13\) in the denominator suggests a potential use for this method as it can be completed to form a perfect square.

Key steps for solving integrals using trigonometric substitution typically involve:
1. Completing the square in the quadratic expression.
2. Choosing an appropriate trigonometric substitution based on the completed square form.
3. Substituting and simplifying the integral.
4. Solving the new integral in terms of the trigonometric variable.
5. Converting back to the original variable \(t\) upon integration.

Each step needs careful consideration to ensure the integration process is correctly executed. It is essential to have a strong foundation in trigonometry and integration techniques to effectively solve this type of problem.
Transcribed Image Text:### Problem Statement **Integrate using the trigonometric substitution method:** \[ \int \frac{3t - 9}{t^2 - 6t + 13} \, dt \] ### Explanation In this problem, our goal is to evaluate the integral of the function \(\frac{3t - 9}{t^2 - 6t + 13}\) with respect to \(t\) by employing the trigonometric substitution method. This technique is often used for integrals involving algebraic expressions that can be transformed into trigonometric forms, making them easier to integrate. The expression \(t^2 - 6t + 13\) in the denominator suggests a potential use for this method as it can be completed to form a perfect square. Key steps for solving integrals using trigonometric substitution typically involve: 1. Completing the square in the quadratic expression. 2. Choosing an appropriate trigonometric substitution based on the completed square form. 3. Substituting and simplifying the integral. 4. Solving the new integral in terms of the trigonometric variable. 5. Converting back to the original variable \(t\) upon integration. Each step needs careful consideration to ensure the integration process is correctly executed. It is essential to have a strong foundation in trigonometry and integration techniques to effectively solve this type of problem.
Expert Solution
Step 1

Given:

3t-9t2-6t+13dt

We have to integrate using the trigonometric substitution method.

 

Step 2

Consider,

3t-9t2-6t+13dt=3(t-3)t2-2×3t+9-9+13dt=3(t-3)(t-3)2+4dt                    t2-2×3t+9=(t-3)2

Now Substitute, t-3=2tan(x) dt=2sec2(x) dx

We get,

3(t-3)(t-3)2+4dt =3  (2 tan(x))(2 tan(x))2+42 sec2(x) dx

steps

Step by step

Solved in 4 steps

Blurred answer
Knowledge Booster
Functions and Inverse Functions
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Precalculus
Precalculus
Calculus
ISBN:
9780135189405
Author:
Michael Sullivan
Publisher:
PEARSON
Calculus: Early Transcendental Functions
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning