Explanation Given Information: The provided substitutions are u = 1 − x 2 , u = x 2 and u = − x 2 , the provided integral is ∫ 1 − x 2 d x . Formula used: Follow the steps as mentioned below to solve an integral ∫ f d x by using substitution rule. Step1: Assume a function u in terms of x from f . Step2: Take the derivative d u d x of the function obtained in step1 and solve for d x in terms of d u . Step3: Substitute the value of u from step 1 and the value of d x from step 2 in the provided integral and then solve the integral Proof: Consider the integral, ∫ 1 − x 2 d x Also consider the substitution, u = 1 − x 2 Take the derivative d u d x of the function u = 1 − x 2 , d u d x = − 2 x Solve for d x , d x = − d u 2 x Now substitute u = 1 − x 2 and d x = − d u 2 x in the integral ∫ 1 − x 2 d x , this gives, ∫ 1 − x 2 d x = ∫ u ⋅ ( − d u 2 x ) = ∫ − u 2 x d u As, the term x is introduced in the denominator

Finite Mathematics and Applied Calculus (MindTap Course List)

7th Edition

ISBN: 9781337274203

Author: Stefan Waner, Steven Costenoble

Publisher: Cengage Learning

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Fundamentals of Trigonometric Identities

Trigonometric Identities are the equalities that involve trigonometry functions and hold for all the values of variables given in the equation.

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Chapter 13.2, Problem 110E

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To prove: That the substitutions u=1−x2, u=x2 and u=−x2 are not useful for the integral ∫1−x2dx.

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Chapter 13.2, Problem 109E

Chapter 13.3, Problem 1E

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Chapter 13 Solutions

Finite Mathematics and Applied Calculus (MindTap Course List)

Ch. 13.1 - In Exercises 142, evaluate the integral. [HinT:...Ch. 13.1 - Prob. 2E Ch. 13.1 - Prob. 3E Ch. 13.1 - Prob. 4E Ch. 13.1 - Prob. 5E Ch. 13.1 - Prob. 6E Ch. 13.1 - Prob. 7E Ch. 13.1 - Prob. 8E Ch. 13.1 - Prob. 9E Ch. 13.1 - Prob. 10E

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To prove: That the substitutions u = 1 − x 2 , u = x 2 and u = − x 2 are not useful for the integral ∫ 1 − x 2 d x .

Solution Summary: The author explains how to solve an integral by using substitution rule.

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To prove: That the substitutions u=1−x2, u=x2 and u=−x2 are not useful for the integral ∫1−x2dx.

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Chapter 13 Solutions