Explanation Given: ∫ π / 4 π / 2 cot 2 x csc 3 x ⋅ d x Key Concept Applied: Reduction Formula: Many trigonometric functions can be integrated by combining substitution and Integration by Parts with the appropriate trigonometric identities. We have to find the value of integral by applying various techniques like reduction formula for sine and cosine, Integration by parts and other techniques. In this problem we find the value of given integral using the formula given in the table. Formula Used: Integration using Reduction formula ∫ csc n u ⋅ d u = − csc n − 2 u ⋅ cot u n − 1 + n − 2 n − 1 ∫ csc n − 2 u ⋅ d u Trigonometric Identity cot 2 x = ( csc 2 x − 1 ) Common Integration Rules ∫ csc 2 u ⋅ d u = − cot u + c ∫ c f ( u ) ⋅ d u = c ⋅ ∫ f ( u ) ⋅ d u ∫ ( f ( u ) ± g ( u ) ) ⋅ d u = ∫ f ( u ) ⋅ d u ± ∫ g ( u ) ⋅ d u ∫ csc x ⋅ d x = ln | tan ( x 2 ) | Calculation: Using Trigonometric Identity to rewrite the given integral ∫ π / 4 π / 2 cot 2 x csc 3 x ⋅ d x ∫ cot 2 x ⋅ csc 3 x ⋅ d x = ∫ ( csc 5 x − csc 3 x ) ⋅ d x = ∫ csc 5 x ⋅ d x − ∫ csc 3 x ⋅ d x Using Trigonometric identity to rewrite: cot 2 x ⋅ csc 3 x = csc 3 x ( csc 2 x − 1 ) = csc 5 x − csc 3 x Using reduction formula to integrate ∫ csc n u ⋅ d u = − csc n − 2 u ⋅ cot u n − 1 + n − 2 n − 1 ∫ csc n − 2 u ⋅ d u ∫ csc 5 x ⋅ d x − ∫ csc 3 x ⋅ d x = ( − csc 5 − 2 x ⋅ cot x 4 + 5 − 2 5 − 1 ∫ csc 5 − 2 x ⋅ d x ) − ( − csc 3 − 2 x ⋅ cot x 2 + 3 − 2 3 − 1 ∫ csc 3 − 2 x ⋅ d x ) = ( − csc 3 x ⋅ cot x 4 + 3 4 ∫ csc 3 x ⋅ d x ) − ( − csc 2 x ⋅ cot x 2 + 1 2 ∫ csc 2 x ⋅ d x ) = (

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 8, Problem 33CRE

To determine

To evaluate:

The given integral

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Chapter 8, Problem 32CRE

Chapter 8, Problem 34CRE

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Chapter 8 Solutions

CALCULUS (CLOTH)

Ch. 8.1 - Prob. 1PQ Ch. 8.1 - Prob. 2PQ Ch. 8.1 - Prob. 3PQ Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E

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The given integral

Solution Summary: The author explains that many trigonometric functions can be integrated by combining substitution and Integration by Parts with the appropriate trigometric identities. They find the value of integral using the formula given in the table.

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To evaluate: The given integral

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Chapter 8 Solutions

To evaluate:

The given integral