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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
8장, 문제 8.1.54e

Using different substitutions
Show that the integral
∫((x² - 1)(x + 1))^(-2/3) dx
can be evaluated with any of the following substitutions.
e. u = tan^(-1) ((x - 1)/2)
What is the value of the integral?

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1
Start with the integral \( \int ((x^2 - 1)(x + 1))^{-2/3} \, dx \). First, simplify the expression inside the integral. Notice that \( x^2 - 1 = (x - 1)(x + 1) \), so the integrand becomes \( ((x - 1)(x + 1)^2)^{-2/3} \).
Rewrite the integrand as \( (x - 1)^{-2/3} (x + 1)^{-4/3} \) by distributing the exponent \( -2/3 \) to each factor inside the parentheses.
Use the substitution \( u = \tan^{-1} \left( \frac{x - 1}{2} \right) \). This means \( \frac{x - 1}{2} = \tan u \), so \( x - 1 = 2 \tan u \).
Differentiate \( x - 1 = 2 \tan u \) with respect to \( u \) to find \( dx \) in terms of \( du \): \( dx = 2 \sec^2 u \, du \).
Express \( (x + 1) \) in terms of \( u \) using the substitution and simplify the integrand accordingly. Then rewrite the entire integral in terms of \( u \) and \( du \), which will allow you to integrate with respect to \( u \).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Integration by Substitution

Integration by substitution is a method to simplify integrals by changing variables. It involves choosing a substitution u = g(x) that transforms the integral into a simpler form in terms of u, making it easier to evaluate. This technique is especially useful when the integrand contains composite functions.
추천 영상:
04:27
Substitution With an Extra Variable

Inverse Trigonometric Substitution

Inverse trigonometric substitution involves substituting a variable with an inverse trigonometric function, such as u = arctan((x - a)/b), to simplify integrals involving algebraic expressions. This substitution leverages trigonometric identities to transform complicated expressions into integrable forms.
추천 영상:
06:35
Derivatives of Other Inverse Trigonometric Functions

Handling Rational Exponents in Integrals

Integrals with rational exponents, like powers of -2/3, require careful manipulation of the integrand. Understanding how to rewrite and simplify expressions with fractional powers is essential, often involving factoring or substitution to convert the integral into a standard form.
추천 영상:
7:39
Introduction to Exponent Rules
관련 실천
교과서 질문

The instructions for the integrals in Exercises 1–10 have three parts, one for the Midpoint Rule, one for the Trapezoidal Rule, and one for Simpson’s Rule.

II. Using the Trapezoidal Rule

b. Evaluate the integral directly and find |ET|.

∫ from 0 to 2 of (t³ + t) dt

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교과서 질문

Using different substitutions

Show that the integral

∫((x² - 1)(x + 1))^(-2/3) dx

can be evaluated with any of the following substitutions.

f. u = arccos x

What is the value of the integral?

31
views
교과서 질문

The instructions for the integrals in Exercises 1–10 have three parts, one for the Midpoint Rule, one for the Trapezoidal Rule, and one for Simpson’s Rule.

II. Using the Trapezoidal Rule

b. Evaluate the integral directly and find |ET|.

∫ from -2 to 0 of (x² - 1) dx

18
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교과서 질문

The instructions for the integrals in Exercises 1–10 have three parts, one for the Midpoint Rule, one for the Trapezoidal Rule, and one for Simpson’s Rule.

II. Using the Trapezoidal Rule

b. Evaluate the integral directly and find |ET|.

∫ from 1 to 3 of (2x - 1) dx

23
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교과서 질문

The instructions for the integrals in Exercises 1–10 have three parts, one for the Midpoint Rule, one for the Trapezoidal Rule, and one for Simpson’s Rule.

II. Using the Trapezoidal Rule

b. Evaluate the integral directly and find |ET|.

∫ from 1 to 2 of 1 / s² ds

21
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교과서 질문

The instructions for the integrals in Exercises 1–10 have three parts, one for the Midpoint Rule, one for the Trapezoidal Rule, and one for Simpson’s Rule.

II. Using the Trapezoidal Rule

b. Evaluate the integral directly and find |ET|.

∫ from 1 to 2 of x dx

29
views