Explanation Formula used: Integration by parts: “Suppose u and v are differentiable functions, then ∫ u d v = u v − ∫ v d u ”. In problem, the value of u = x and d v = cos x d x . Integrate the term d v = cos x d x to obtain the value of v . v = ∫ cos x d x = sin x Thus, the value of v = sin x . The derivative of u = x is d u = d x

Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

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ISBN: 9780134996103

Author: William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz

Publisher: PEARSON

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Chapter 8.2, Problem 2E

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To evaluate: The value of ∫xcosxdx with u=x and dv=cosxdx by using integration by parts.

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Chapter 8.2, Problem 1E

Chapter 8.2, Problem 3E

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Chapter 8 Solutions

Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

Ch. 8.1 - What change of variable would you use for the...Ch. 8.1 - Prob. 2QC Ch. 8.1 - Prob. 3QC Ch. 8.1 - Prob. 4QC Ch. 8.1 - What change of variables would you use for the...Ch. 8.1 - Prob. 2E Ch. 8.1 - What trigonometric identity is useful in...Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E

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The value of ∫ x cos x d x with u = x and d v = cos x d x by using integration by parts.

Solution Summary: The author explains how to evaluate the value of displaystyleint xmathrmcosxdx using integration by parts.

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To evaluate: The value of ∫xcosxdx with u=x and dv=cosxdx by using integration by parts.

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Chapter 8 Solutions