Explanation Given: The definite integral ∫ − π π x sin n x d x = { 2 π n , n is odd − 2 π n , n is even . Formula Used: The integration by parts is ∫ u d v = u v − ∫ v d u . Proof: Consider the integral, ∫ − π π x sin n x d x Use integration by parts. Let, u = x Then, d u = d x And, d v = sin n x d x Then, v = ∫ sin n x d x = − 1 n cos n x Therefore, ∫ − π π x sin n x d x = [ − x n cos n x − ∫ ( − 1 n cos n x ) d x ] − π π = [ − x n cos n x + 1 n ∫ ( cos n x ) d x ] − π π = [ − x n cos n x + 1 n 2 sin n x ] − π π =

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ISBN: 9781285948133

Author: Ron Larson; Bruce H. Edwards

Publisher: Cengage Learning US

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Chapter 8.2, Problem 91E

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To prove: That the definite integral gives ∫−ππxsinnxdx={ 2πn,n is odd−2πn,n is even for the given condition.

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Chapter 8.2, Problem 90E

Chapter 8.2, Problem 92E

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To prove: That the definite integral gives ∫ − π π x sin n x d x = { 2 π n , n is odd − 2 π n , n is even for the given condition.

Solution Summary: The author explains that the definite integral gives displaystyle 'underset'-pi' overset, and the integration by parts.

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To prove: That the definite integral gives ∫−ππxsinnxdx={ 2πn,n is odd−2πn,n is even for the given condition.

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Chapter 8 Solutions