Integrals are not always easy to evaluate; sometimes we need to be clever! In this problem we study the definite integral I=∫(π>0) xsin(x)/1+cos^2(x) dx. At first, it appears difficult to evaluate this integral with substitution (you can try with simple substitution ideas...). But it can be done! Let's see how it goes. (a) Use the substitution u=π−x to show that I= π/2∫(π>0)sin(x)/1+cos^2(x) dx. (Recall that sin(π−x)=sin(x) and cos(π−x)=−cos(x).)
Integrals are not always easy to evaluate; sometimes we need to be clever! In this problem we study the definite integral I=∫(π>0) xsin(x)/1+cos^2(x) dx. At first, it appears difficult to evaluate this integral with substitution (you can try with simple substitution ideas...). But it can be done! Let's see how it goes. (a) Use the substitution u=π−x to show that I= π/2∫(π>0)sin(x)/1+cos^2(x) dx. (Recall that sin(π−x)=sin(x) and cos(π−x)=−cos(x).)
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Integrals are not always easy to evaluate; sometimes we need to be clever! In this problem we study the definite
I=∫(π>0) xsin(x)/1+cos^2(x) dx.
At first, it appears difficult to evaluate this integral with substitution (you can try with simple substitution ideas...). But it can be done! Let's see how it goes.
(a) Use the substitution u=π−x to show that
I= π/2∫(π>0)sin(x)/1+cos^2(x) dx.
(Recall that sin(π−x)=sin(x) and cos(π−x)=−cos(x).)
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