Explanation Given Information: The equation is ∫ csc x d x = − ln | csc x + cot x | + C . Formula Used: The following integration formula is used: ∫ 1 x d x = ln | x | + C The derivation formula, d d x ( csc x ) = − csc x cot x and d d x cot x = − csc 2 x Proof: Consider the left side function of the provided equation, ∫ csc x d x . Multiply and divide the function by ( csc x + cot x ) . ∫ csc x d x = ∫ csc x csc x + cot x csc x + cot x d x Replace the parenthetical expression with variable u . u = csc x + cot x Determine the derivative of u with respect to x. d u d x = d d x ( csc x + cot x ) d u d x = − csc x cot x − csc 2 x Determine the value of d x

Student Solutions Manual for Waner/Costenoble's Finite Math and Applied Calculus, 7th

7th Edition

ISBN: 9781337275972

Author: Waner, Stefan; Costenoble, Steven

Publisher: Cengage Learning

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Chapter 16.3, Problem 32E

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To prove: The derivation of the function ∫cscx dx=−ln|cscx+cotx|+C where a, b are constants with a≠0.

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Chapter 16.3, Problem 31E

Chapter 16.3, Problem 33E

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Chapter 16 Solutions

Student Solutions Manual for Waner/Costenoble's Finite Math and Applied Calculus, 7th

Ch. 16.1 - Prob. 1E Ch. 16.1 - Prob. 2E Ch. 16.1 - Prob. 3E Ch. 16.1 - Prob. 4E Ch. 16.1 - Prob. 5E Ch. 16.1 - Prob. 6E Ch. 16.1 - Prob. 7E Ch. 16.1 - Prob. 8E Ch. 16.1 - Prob. 9E Ch. 16.1 - Prob. 10E

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To prove: The derivation of the function ∫ csc x d x = − ln | csc x + cot x | + C where a, b are constants with a ≠ 0 .

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To prove: The derivation of the function ∫cscx dx=−ln|cscx+cotx|+C where a, b are constants with a≠0.

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Chapter 16 Solutions