Explanation Given information: Use of identity sin 2 x = 1 2 ( 1 − cos 2 x ) & cos 2 x = 1 2 ( 1 + cos 2 x ) Concept and Formula Used: We convert higher power of sine and cosine function using the identity sin 2 x = 1 2 ( 1 − cos 2 x ) , cos 2 x = 1 2 ( 1 + cos 2 x ) , cos A cos B = 1 2 [ cos ( a + b ) + cos ( a − b ) ] , ∫ cos x = sin x & ∫ 1 d x = x Calculation: It is given that ∫ sin 4 x cos 2 x d x = ∫ ( sin 2 x ) 2 cos 2 x d x , Using sin 2 x = 1 2 ( 1 − cos 2 x ) & cos 2 x = 1 2 ( 1 + cos 2 x ) ∫ ( 1 2 ( 1 − cos 2 x ) ) 2 ( 1 2 ( 1 + cos 2 x ) ) d x = ∫ 1 8 ( 1 + cos 2 2 x − 2 cos 2 x ) ( 1 + cos 2 x ) d x Using cos 2 x = 1 2 ( 1 + cos 2 x ) ∫ sin 4 x cos 2 x d x = 1 8 ∫ ( 1 + ( 1 + cos 4 x 2 ) − 2 cos 2 x ) ( 1 + cos 2 x ) d x = 1 8 ∫ ( 1 + 1 2 + 1 2 cos 4 x − 2 cos 2 x ) ( 1 + cos 2 x ) d x = 1 8 ∫ ( 3 2 + 1 2 cos 4 x − 2 cos 2 x ) ( 1 + cos 2 x ) d x = 1 8 ∫ ( 3 2 + 3 2 cos 2 x + 1 2 cos 4 x + 1 2 cos 4 x cos 2 x − 2 cos 2 x − 2 cos 2

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 8.2, Problem 69E

To determine

To calculate:

The integral

∫sin4xcos2x dx.

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Chapter 8.2, Problem 68E

Chapter 8.2, Problem 70E

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Chapter 8 Solutions

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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 integral ∫ sin 4 x cos 2 x d x .

Solution Summary: The author explains how to convert higher power of sine and cosine function using the identity lmathrmsin

Videos

To calculate:

The integral

∫sin4xcos2x dx.

Want to see the full answer?

Chapter 8 Solutions

The integral ∫ sin 4 x cos 2 x d x .

Solution Summary: The author explains how to convert higher power of sine and cosine function using the identity lmathrmsin

Videos

To calculate: The integral ∫sin4xcos2x dx.

Want to see the full answer?

Chapter 8 Solutions

To calculate:

The integral

∫sin4xcos2x dx.