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?

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Start with the integral \( \int ((x^{2} - 1)(x + 1))^{-\frac{2}{3}} \, dx \). First, simplify the expression inside the integral. Notice that \( x^{2} - 1 = (x - 1)(x + 1) \), so the integrand becomes \( ((x - 1)(x + 1)(x + 1))^{-\frac{2}{3}} = ((x - 1)(x + 1)^{2})^{-\frac{2}{3}} \).

Rewrite the integrand as \( (x - 1)^{-\frac{2}{3}} (x + 1)^{-\frac{4}{3}} \) by distributing the exponent \( -\frac{2}{3} \) to each factor.

Use the substitution \( u = \arccos x \). Then, \( x = \cos u \) and \( dx = -\sin u \, du \). This substitution is useful because it relates \( x \) to trigonometric functions, which can simplify expressions involving \( x^{2} - 1 \).

Express the factors \( x - 1 \) and \( x + 1 \) in terms of \( u \): \( x - 1 = \cos u - 1 \) and \( x + 1 = \cos u + 1 \). Also, note that \( \sin^{2} u = 1 - \cos^{2} u \), which can help simplify the expression further.

Rewrite the integral entirely in terms of \( u \) and \( du \), substituting \( x \), \( dx \), and the factors \( x - 1 \), \( x + 1 \). Then simplify the resulting integral, which should be easier to evaluate using standard trigonometric integral techniques.

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주요 개념

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Trigonometric Substitution

Trigonometric substitution is a technique used to simplify integrals involving algebraic expressions by substituting a trigonometric function for a variable. For example, substituting u = arccos(x) transforms expressions involving x into trigonometric forms, often making the integral easier to evaluate by leveraging trigonometric identities.

Integration of Powers of Functions

Integrating expressions raised to fractional powers, such as ((x² - 1)(x + 1))^(-2/3), requires understanding how to manipulate and simplify the integrand. This often involves rewriting the expression in a more manageable form or using substitution to convert it into a standard integral form.

Inverse Trigonometric Functions and Their Derivatives

Inverse trigonometric functions like arccos(x) have specific derivatives that are essential when performing substitution in integrals. Knowing that d/dx[arccos(x)] = -1/√(1 - x²) helps in changing variables and adjusting the differential dx accordingly during integration.

Using different substitutionsShow that the integral∫((x² - 1)(x + 1))^(-2/3) dxcan be evaluated with any of the following substitutions.f. u = arccos xWhat is the value of the integral?