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?

Just do it again on the second integral...or use an integral table (they're always handy).

Indeed. [imath]\int x\sin x \ dx[/imath] can be done using integration by parts also.

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?

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.

Oh, cheers for that