Explanation Given: The given integral is ∫ x 2 ( x 2 − 25 ) 3 2 d x . Formula used: ∫ sec n u d u = sec n − 2 ( u ) tan ( u ) n − 1 + n − 2 n − 1 ∫ sec n − 2 ( u ) d u Calculation: Consider integral I = ∫ x 2 ( x 2 − 25 ) 3 2 d x . Write x 2 as x 2 + 25 − 25 . So, I = ∫ ( x 2 + 25 − 25 ) ( x 2 − 25 ) 3 2 d x = ∫ ( x 2 − 25 ) 5 2 d x + ∫ 25 ( x 2 − 25 ) 3 2 d x = ∫ ( x 2 − 25 ) 5 2 d x + 25 ∫ ( x 2 − 25 ) 3 2 d x Substitute, x = 5 sec u → u = sec − 1 ( x 5 ) , d x = 5 sec u tan u d u . I = ∫ ( 25 sec 2 u − 25 ) 5 2 × 5 sec u tan u d u + 25 ∫ ( 25 sec 2 u − 25 ) 3 2 5 sec u tan u d u = ∫ 5 ( 25 sec 2 u − 25 ) 5 2 sec u tan u d u + 15625 ∫ ( sec 2 u − 1 ) 3 2 sec u tan u d u = 15625 ∫ sec u tan 6 u d u + 15625 ∫ sec u tan 4 u d u = 15625 ∫ sec u ( sec 2 u − 1 ) 3 d u + 15625 ∫ sec u ( sec 2 u − 1 ) 2 d u Simplifying further, we get, I = 15625 ∫ sec u ( sec 2 u − 1 ) 3 d u + 15625 ∫ sec u ( sec 2 u − 1 ) 2 d u Expanding the identity, we get, I = 15625 ∫ ( sec 7 u − 3 sec 5 u + 3 sec 3 u − sec u ) d u + 15625 ∫ ( sec 5 u − 2 sec 3 u + sec u ) d u = 15625 ∫ ( sec 7 u − 2 sec 5 u + sec 3 u ) d u = 15625 ( ∫ sec 7 u d u − ∫ 2 sec 5 u d u + ∫ sec 3 u d u ) Applying the reduction formula, we get, ∫ sec n u d u = sec n − 2 ( u ) tan ( u ) n − 1 + n − 2 n − 1 ∫ sec n − 2 ( u ) d u Put, n = 7 ∫ sec 7 u d u = sec 7 − 2 ( u ) tan ( u ) 7 − 1 + 7 − 2 7 − 1 ∫ sec 7 − 2 ( u ) d u ∫ sec 7 u d u = sec 5 ( u ) tan ( u ) 6 + 5 6 ∫ sec 5 ( u ) d u Hence, I = 15625 ( sec 5 ( u ) tan ( u ) 6 + 5 6 ∫ sec 5 u</

Calculus: Early Transcendental Functions

6th Edition

ISBN: 9781305005303

Author: Ron Larson, Bruce Edwards

Publisher: CENGAGE C

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Chapter 8.4, Problem 4E

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To calculate: The definite integral ∫x2(x2−25)32dx by using the trigonometry substitution.

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Chapter 8.4, Problem 3E

Chapter 8.4, Problem 5E

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Find all values x = a where the function is discontinuous. For each value of x, give the limit of the function as x approaches a. Be sure to note when the limit doesn't exist. 8+x f(x) = x(x-1) (Use a comma to separate answers as needed.) OA. The function f is discontinuous at the single value x = OB. The function f is discontinuous at the single value x = OC. The function f is discontinuous at the two values x = OD. The function f is discontinuous at the two values x = not oo or -0. OE. The function f is discontinuous at the two values x = The limit is The limit does not exist and is not oo or - co. The limits for both values do not exist and are not co or - co. The limit for the smaller value is The limit for the larger value does not exist and is The limit for the smaller value does not exist and is not co or - co. The limit for the larger

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Chapter 8 Solutions

Calculus: Early Transcendental Functions

Ch. 8.1 - Choosing an Antiderivative In Exercises 3 and 4,...Ch. 8.1 - Choosing an Antiderivative In Exercises 14, select...Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Choosing a Formula In Exercises 5-14, select the...Ch. 8.1 - Choosing a Formula In Exercises 5-14, select the...Ch. 8.1 - Choosing a Formula In Exercises 5-14, select the...Ch. 8.1 - Choosing a Formula In Exercises 5-14, select the...Ch. 8.1 - Choosing a Formula In Exercises 5-14, select the...Ch. 8.1 - Choosing a Formula In Exercises 5-14, selectthe...

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The definite integral ∫ x 2 ( x 2 − 25 ) 3 2 d x by using the trigonometry substitution.

Solution Summary: The author calculates the definite integral by using the trigonometry substitution. The solution is underset_46875mathrmlnleft.

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To calculate: The definite integral ∫x2(x2−25)32dx by using the trigonometry substitution.

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Chapter 8 Solutions