Explanation Proof: Use the reduction formulas to show that if n is an odd positive integer then ∫ 0 π / 2 sin n x d x = ∫ 0 π / 2 cos n x d x = [ 2 ⋅ 4 ⋅ 6 … … ( n − 1 ) 3 ⋅ 5 ⋅ 7 … … n ] Derive the formula. Using reduction formula, ∫ sin n x d x = n − 1 n ∫ sin n − 2 x d x Now, if n is odd, say n = 2 m + 1 ∫ sin 2 m + 1 x d x = 2 m 2 m + 1 ∫ sin 2 m − 1 x d x Now, apply the same formula then ∫ sin 2 m − 1 x d x = 2 m − 2 2 m − 1 ∫ sin 2 m − 3 x d x So, the original formula becomes ∫ sin 2 m + 1 x d x = 2 m 2 m + 1 ⋅ 2 m − 2 2 m − 1 ∫ sin 2 m − 3 x d x Continue the same way, ∫ sin 2 m + 1 x d x = 2 m 2 m + 1 ⋅ 2 m − 2 2 m − 1 … 2 3 ∫ sin x d x = 2 m 2 m + 1 ⋅ 2 m − 2 2 m − 1 … 2 3 [ − cos x ] Now, use the limits of integration

6th Edition

ISBN: 9781465208880

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 7.2, Problem 60PS

To determine

To prove: the Wallis’s formula

∫0 π/2 sin nxdx=∫0 π/2 cos nxdx=[2⋅4⋅6……(n−1)3⋅5⋅7……n] when n is odd

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Chapter 7.2, Problem 59PS

Chapter 7.3, Problem 1PS

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Chapter 7 Solutions

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To prove: the Wallis’s formula ∫ 0 π / 2 sin n x d x = ∫ 0 π / 2 cos n x d x = [ 2 ⋅ 4 ⋅ 6 … … ( n − 1 ) 3 ⋅ 5 ⋅ 7 … … n ] when n is odd

Solution Summary: The author explains how to prove the Wallis’s formula ldisplaystyle

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To prove: the Wallis’s formula

∫0 π/2 sin nxdx=∫0 π/2 cos nxdx=[2⋅4⋅6……(n−1)3⋅5⋅7……n] when n is odd

Want to see the full answer?

Chapter 7 Solutions

To prove: the Wallis’s formula ∫ 0 π / 2 sin n x d x = ∫ 0 π / 2 cos n x d x = [ 2 ⋅ 4 ⋅ 6 … … ( n − 1 ) 3 ⋅ 5 ⋅ 7 … … n ] when n is odd

Solution Summary: The author explains how to prove the Wallis’s formula ldisplaystyle

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To prove: the Wallis’s formula ∫0 π/2 sin nxdx=∫0 π/2 cos nxdx=[2⋅4⋅6……(n−1)3⋅5⋅7……n] when n is odd

Want to see the full answer?

Chapter 7 Solutions

To prove: the Wallis’s formula

∫0 π/2 sin nxdx=∫0 π/2 cos nxdx=[2⋅4⋅6……(n−1)3⋅5⋅7……n] when n is odd