Explanation Given information: The term C n is given as, C n = ∫ − π 2 π 2 cos n θ d θ Proof: Consider the term cos n θ . This can be written as cos θ × cos n − 1 θ . Then, the integration should be, C n = ∫ − π 2 π 2 c o s θ cos n − 1 θ d θ Now, there are two parts of the integral and considering these two parts we can perform the integration by parts. C n = ∫ − π 2 π 2 c o s θ cos n − 1 θ d θ = ∫ − π 2 π 2 d ( sin θ ) d θ cos n − 1 θ d θ = [ sin θ cos n − 1 θ ] π 2 − π 2 − ∫ − π 2 π 2 sin θ d ( cos n − 1 θ ) d θ d θ = [ sin θ cos n − 1 θ ] π 2 − π 2 − ∫ − π 2 π 2 sin θ ( n − 1 ) cos n − 2 θ ( − sin θ ) d θ = [ sin θ cos n − 1 θ ] π 2 − π 2 + ( n − 1 ) ∫ − π 2 π 2 sin 2 θ cos n − 2 θ d θ = ( sin ( π 2 ) cos n − 1 ( π 2 ) − sin ( − π 2 ) cos n − 1 ( − π 2 ) ) + ( n − 1 ) ∫ − π 2 π 2 sin 2 θ cos n − 2 θ d θ = 0 + ( n − 1 ) ∫ − π 2 π 2 sin 2 θ cos n − 2 θ d θ C n = ( n − 1 ) ∫ − π 2 π 2 sin 2 θ cos n − 2 θ d θ Because sin 2 θ + cos 2 θ = 1 , we can substitute sin 2 θ = 1 − cos 2 θ int the obtained C n

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Question

Chapter 16.3, Problem 43E

To determine

To prove:

That Cn=(n−1n)Cn−2 using the method of integration by parts.

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Chapter 16.3, Problem 42E

Chapter 16.3, Problem 44E

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Chapter 16 Solutions

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To prove: That C n = ( n − 1 n ) C n − 2 using the method of integration by parts.

Solution Summary: The author explains how to perform the integration by parts using the term C_n.

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To prove:

That Cn=(n−1n)Cn−2 using the method of integration by parts.

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Chapter 16 Solutions

To prove: That C n = ( n − 1 n ) C n − 2 using the method of integration by parts.

Solution Summary: The author explains how to perform the integration by parts using the term C_n.

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To prove: That Cn=(n−1n)Cn−2 using the method of integration by parts.

Want to see the full answer?

Chapter 16 Solutions

To prove:

That Cn=(n−1n)Cn−2 using the method of integration by parts.