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 8.6.56
Chapter 8, Problem 8.6.56

Evaluate the integrals in Exercises 51–56 by making a substitution (possibly trigonometric) and then applying a reduction formula.
∫ (from 0 to 1/√3) dt / (t² + 1)^(7/2)

Verified step by step guidance
1
Recognize that the integral has the form \(\int \frac{dt}{(t^2 + 1)^{7/2}}\), which suggests a trigonometric substitution because of the \(t^2 + 1\) term inside the power.
Use the substitution \(t = \tan(\theta)\), which implies \(dt = \sec^2(\theta) d\theta\) and \(t^2 + 1 = \sec^2(\theta)\).
Rewrite the integral in terms of \(\theta\): replace \(dt\) with \(\sec^2(\theta) d\theta\) and \((t^2 + 1)^{7/2}\) with \((\sec^2(\theta))^{7/2} = \sec^7(\theta)\), so the integrand becomes \(\frac{\sec^2(\theta)}{\sec^7(\theta)} = \sec^{-5}(\theta)\).
Change the limits of integration from \(t\) to \(\theta\): when \(t=0\), \(\theta = \arctan(0) = 0\); when \(t=\frac{1}{\sqrt{3}}\), \(\theta = \arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}\).
Now the integral is \(\int_0^{\pi/6} \sec^{-5}(\theta) d\theta = \int_0^{\pi/6} \cos^5(\theta) d\theta\). Use a reduction formula for powers of cosine to evaluate this integral step-by-step.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
9m
Was this helpful?

Key Concepts

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

Trigonometric Substitution

Trigonometric substitution is a technique used to simplify integrals involving expressions like √(a² + x²), √(a² - x²), or √(x² - a²). By substituting x with a trigonometric function (e.g., x = a tan θ), the integral transforms into a trigonometric integral that is often easier to evaluate.
Recommended video:
6:04
Introduction to Trigonometric Functions

Reduction Formulas

Reduction formulas express an integral with a certain power or parameter in terms of a similar integral with a lower power or simpler parameter. They help solve complex integrals step-by-step by breaking them down into simpler, more manageable integrals.
Recommended video:
Guided course
5:59
Recursive Formulas

Definite Integration with Limits

Definite integration involves evaluating the integral between specified limits. When using substitution, it is important to change the limits according to the substitution or revert to the original variable before applying the limits to find the exact numerical value.
Recommended video:
05:43
Definition of the Definite Integral
Related Practice
Textbook Question

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

∫ x (7x + 5)^(3/2) dx

Textbook Question

Evaluate the integrals in Exercises 1–24 using integration by parts.

∫ arcsin(y) dy

Textbook Question

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.

∫ (√x / (1 + x³)) dx

Hint: Let u = x^(3/2).

Textbook Question

In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer.

∫ from 1 to ∞ of ((1 / (e^x - 2^x)) dx)

1
views
Textbook Question

For Exercises 49–52, complete the square before using an appropriate trigonometric substitution.

∫ √(x² + 2x + 2) / (x² + 2x + 1) dx

Textbook Question

[Technology Exercise] When solving Exercises 33-40, you may need to use a calculator or a computer.

Find, to two decimal places, the areas of the surfaces generated by revolving the curves in Exercises 35 and 36 about the x-axis.

y = sin x, 0 ≤ x ≤ π

1
views