Explanation Formula used: ∫ x n d x = x n + 1 n + 1 + C d d x ( sec a x ) = a sec a x tan a x sec 2 a x − tan 2 a x = 1 ∫ sec x d x = ln | sec x + tan x | + C Calculation: Obtain the value of the integral by using substitution method, Let x = 5 3 sec θ . Then d x = 5 3 sec θ tan θ d θ . Therefore, the integral becomes, ∫ d x 9 x 2 − 25 = ∫ 5 3 sec θ tan θ 9 ( 5 3 sec θ ) 2 − 25 d θ = 5 3 ∫ sec θ tan θ 9 ( 25 9 sec 2 θ ) − 25 d θ = 5 3 ∫ sec θ tan θ 25 sec 2 θ − 25 d θ = ( 5 3 × 1 5 ) ∫ sec θ tan θ sec 2 θ − 1 d θ On further simplification, ∫ d x 9 x 2 − 25 = 1 3 ∫ sec θ tan θ tan

Student Solutions Manual, Single Variable for Calculus: Early Transcendentals

2nd Edition

ISBN: 9780321954329

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

Publisher: PEARSON

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Chapter 7, Problem 50RE

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To evaluate: The integral ∫dx9x2−25, where x>53.

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Chapter 7, Problem 49RE

Chapter 7, Problem 51RE

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Chapter 7 Solutions

Student Solutions Manual, Single Variable for Calculus: Early Transcendentals

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