Crash Posted June 1, 2005 Share Posted June 1, 2005 Ive got the question S x^2cos(x) dx But by using S u(x)v'(x)= u(x)v(x)-S u'(x)v(x) dx So using v(x)= Sin(x) and u(x)= x^2 Im left with x^2Sin(x)- S 2xsin(x) dx with the end part being a product also can someone help me? Link to comment Share on other sites More sharing options...
fuhrerkeebs Posted June 1, 2005 Share Posted June 1, 2005 Just do it again on the second integral...or use an integral table (they're always handy). Link to comment Share on other sites More sharing options...
Dave Posted June 1, 2005 Share Posted June 1, 2005 Indeed. [imath]\int x\sin x \ dx[/imath] can be done using integration by parts also. Link to comment Share on other sites More sharing options...
Crash Posted June 1, 2005 Author Share Posted June 1, 2005 ok so i got for my result =x^2sin(x) + 2xcos(x) - 2sin(x) + c But when i derive this again to check the awnser i have a wrong sign somewhere and i cant find where and i been through over and over and over.... Is this the right intergral? Link to comment Share on other sites More sharing options...
Ducky Havok Posted June 1, 2005 Share Posted June 1, 2005 It's right. The derivative of what you typed is [math]2xsin(x)+x^2cos(x)+2cos(x)-2xsin(x)-2cos(x)[/math] Then everything but [math]x^{2}cos(x)[/math] cancles out. Link to comment Share on other sites More sharing options...
Crash Posted June 3, 2005 Author Share Posted June 3, 2005 Oh, cheers for that Link to comment Share on other sites More sharing options...
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