October 27th, 2010, 06:15 PM  #1 
Senior Member Joined: Oct 2010 Posts: 126 Thanks: 0  Integration
Hi, I require help with this problem. I'm asking kindly for help. Hints... . 
October 27th, 2010, 07:29 PM  #2  
Senior Member Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14  Re: Integration Quote:
Therefor.... So,  
October 27th, 2010, 07:51 PM  #3 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs  Re: Integration
We have: A definite integral can be evaluated using integration by parts in the following manner: We will let and , thus and Putting it together, we have: The expression to the left of the integral is zero, and using the Pythagorean identity for , we have: Solve for Thus, edit: Sorry, stainburg, I didn't catch that you had already solved. 
October 27th, 2010, 07:55 PM  #4  
Senior Member Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14  Re: Integration Quote:
 
October 27th, 2010, 08:04 PM  #5 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs  Re: Integration
I require very little sleep. 
October 27th, 2010, 08:15 PM  #6  
Senior Member Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14  Re: Integration Quote:
 
November 2nd, 2010, 07:06 PM  #7 
Senior Member Joined: Oct 2010 Posts: 126 Thanks: 0  Re: Integration
Thanks. The next part may or not be a followup to the question. I worked it out. I just think it's unnecessarily tedious. Hmmm...ponders Then for the definite integral: 
November 2nd, 2010, 07:42 PM  #8 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs  Re: Integration
You obtained the correct result (according to my TI89) and in a much more elegant form. As a minor aside, you don't write the constant C after the result from a definite integral...it essentially gets subtracted away when you apply the F(b)  F(a) part of the antiderivative form of the fundamental theorem of calculus...i.e. [F(b) + C]  [F(a) + C] = F(b)  F(a) 
November 3rd, 2010, 01:26 AM  #9  
Senior Member Joined: Oct 2010 Posts: 126 Thanks: 0  Re: Integration Quote:
 

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