Skip to main content
Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 8 - Techniques of IntegrationProblem 8.1.28
Chapter 8, Problem 8.1.28

The integrals in Exercises 1–44 are in no particular order. Evaluate each integral using any algebraic method, trigonometric identity, or substitution you think is appropriate.
∫ (dx / ((x - 2)√(x² - 4x + 3)))

Verified step by step guidance
1
Start by simplifying the expression under the square root. Notice that \(x^2 - 4x + 3\) can be factored or completed the square. Rewrite it as \(x^2 - 4x + 3 = (x - 2)^2 - 1\).
Make the substitution \(u = x - 2\) to simplify the integral. Then, \(du = dx\), and the integral becomes \(\int \frac{du}{u \sqrt{u^2 - 1}}\).
Recognize that the integral now has the form \(\int \frac{du}{u \sqrt{u^2 - 1}}\), which suggests using a trigonometric substitution. Since \(\sqrt{u^2 - 1}\) appears, consider the substitution \(u = \sec \theta\), so that \(\sqrt{u^2 - 1} = \tan \theta\).
Express \(du\) in terms of \(d\theta\): since \(u = \sec \theta\), then \(du = \sec \theta \tan \theta \, d\theta\). Substitute these into the integral to rewrite it entirely in terms of \(\theta\).
Simplify the resulting integral in \(\theta\) and integrate using standard trigonometric integral formulas. After integration, substitute back \(\theta = \sec^{-1}(u)\) and then \(u = x - 2\) to express the answer in terms of \(x\).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
2m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Integration by Substitution

Integration by substitution involves changing variables to simplify an integral. By letting a new variable represent a part of the integrand, the integral can often be transformed into a more manageable form. This technique is especially useful when the integrand contains composite functions or expressions under radicals.
Recommended video:
04:27
Substitution With an Extra Variable

Simplifying Radicals and Quadratic Expressions

Simplifying the expression under the square root, often a quadratic, helps in recognizing patterns or suitable substitutions. Completing the square or factoring the quadratic can transform the radical into a standard form, making the integral easier to evaluate using known formulas or trigonometric substitutions.
Recommended video:
6:36
Simplifying Trig Expressions

Trigonometric Substitution

Trigonometric substitution replaces algebraic expressions involving square roots of quadratic polynomials with trigonometric functions. This method leverages identities like sin²θ + cos²θ = 1 to simplify integrals containing radicals, converting them into integrals of trigonometric functions that are easier to solve.
Recommended video:
6:04
Introduction to Trigonometric Functions
Related Practice
Textbook Question

In Exercises 69–80, determine whether the improper integral converges or diverges. If it converges, evaluate the integral.

∫₋∞⁰ x² e^(x³) dx

1
views
Textbook Question

In Exercises 17–20, express the integrand as a sum of partial fractions and evaluate the integrals.

∫ (x³ dx) / (x² - 2x + 1) from -1 to 0

1
views
Textbook Question

Use the table of integrals at the back of the text to evaluate the integrals in Exercises 1–26.

∫ √(x² - 4) / x dx

Textbook Question

The length of one arch of the curve y = sin x is given by

L = ∫(from 0 to π) √(1 + cos²(x)) dx.

Estimate L by Simpson's Rule with n = 8.

Textbook Question

In Exercises 39–48, use an appropriate substitution and then a trigonometric substitution to evaluate the integrals.

∫ (e^{t} dt) / ((1 + e^{2t})^{3/2}) from ln(3/4) to ln(4/3)

1
views
Textbook Question

Solve the initial value problems in Exercises 53–56 for y as a function of x.

(x² + 1)² (dy/dx) = √(x² + 1), where y(0) = 1