Question 395544
Let {{{u = 2x - 3}}}, {{{du = 2dx}}}. We now have


{{{int((sin(2011u*pi)*u^2011), du)}}} (I'm writing it at an indefinite integral for now. Don't forget the 1/2 at the end)


Letting {{{y = u^2011}}}, {{{dz = sin (2011u*pi) du}}}, we obtain {{{dy = 2011u^2010du}}} and {{{z = (1/(2011*pi))*(-cos (2011u*pi))}}}


Applying integration by parts,


{{{int((sin(2011u*pi)*u^2011), du) = (u^2011)(1/(2011*pi))*(-cos(2011u*pi)) -   int ( ( (1/(2011*pi))*(-cos(2011u*pi))*2011u^2010), du)}}}


Which is equal to:

{{{(u^2011)(1/(2011*pi))*(-cos(2011u*pi)) - (1/pi)int ((-cos(2011u*pi)*u^2010), du)}}}


Hopefully, by repeatedly using integration by parts, you should be able to get a sum of 2010 or 2011 terms that should be simplifiable in some way. However the easiest way is obviously to use the nint() function or the definite integral function on a graphing calculator.