Sep 2, 2014 #1 R reza1001gh Newbie level 4 Joined Feb 3, 2012 Messages 6 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,281 Activity points 1,305 How can I calculate the following integral?? x0 , T are constants Please help, thank you for your help in advance Last edited by a moderator: Sep 2, 2014
How can I calculate the following integral?? x0 , T are constants Please help, thank you for your help in advance
Sep 2, 2014 #2 A albbg Advanced Member level 4 Joined Nov 7, 2009 Messages 1,311 Helped 448 Reputation 898 Reaction score 409 Trophy points 1,363 Location Italy Activity points 9,986 reza1001gh, I can't see any text associated to this thread. It seems empty
Sep 2, 2014 #3 FvM Super Moderator Staff member Joined Jan 22, 2008 Messages 52,487 Helped 14,756 Reputation 29,794 Reaction score 14,121 Trophy points 1,393 Location Bochum, Germany Activity points 298,375 A board software problem occured when the OP edited his post.
Sep 2, 2014 #4 Dominik Przyborowski Advanced Member level 4 Joined Jun 6, 2013 Messages 1,211 Helped 505 Reputation 1,015 Reaction score 506 Trophy points 1,393 Location Norway Activity points 9,329 Calculate it for n=1,2 and 3 and try to find recursive formula.
Sep 2, 2014 #5 andre_luis Super Moderator Staff member Joined Nov 7, 2006 Messages 9,595 Helped 1,190 Reputation 2,399 Reaction score 1,208 Trophy points 1,403 Location Brazil Activity points 55,686 Parameters T and n are constant numbers, so that you should integrate something in the form of \[{ sin(x)}^{4n}\]. Check this pattern here:
Parameters T and n are constant numbers, so that you should integrate something in the form of \[{ sin(x)}^{4n}\]. Check this pattern here:
Sep 2, 2014 #6 A albbg Advanced Member level 4 Joined Nov 7, 2009 Messages 1,311 Helped 448 Reputation 898 Reaction score 409 Trophy points 1,363 Location Italy Activity points 9,986 We know that sin(a)*sin(b) = cos(a-b)/2 - cos(a+b)/2 if a=(x+xo)*T and b=(x-xo)*T then: sin[(x+xo)*T]*sin[(x-xo)*T] = cos(2*xo*T)/2-cos(2*x*T)/2 the first term doesn't depends from "x" so it is costant. calling it "k" we will have: sin[(x+xo)*T]*sin[(x-xo)*T] = [k-cos(2*x*T)]/2 so we will have to integrate: {[k-cos(2*x*T)]/2}^(2*n) if "n" is an integer number, the integral can be calculated using the reduction formula (https://en.wikipedia.org/wiki/Integration_by_reduction_formulae). It is quite tedious.
We know that sin(a)*sin(b) = cos(a-b)/2 - cos(a+b)/2 if a=(x+xo)*T and b=(x-xo)*T then: sin[(x+xo)*T]*sin[(x-xo)*T] = cos(2*xo*T)/2-cos(2*x*T)/2 the first term doesn't depends from "x" so it is costant. calling it "k" we will have: sin[(x+xo)*T]*sin[(x-xo)*T] = [k-cos(2*x*T)]/2 so we will have to integrate: {[k-cos(2*x*T)]/2}^(2*n) if "n" is an integer number, the integral can be calculated using the reduction formula (https://en.wikipedia.org/wiki/Integration_by_reduction_formulae). It is quite tedious.