Let, #I=int(5x^3+3x-1)cos^2(2x+1)pidx#.
Since, #cos^2(2x+1)pi=cos^2(2pix+pi)=cos^2(2pix)#,
#=(1+cos(2xx2pix))/2=(1+cos4pix)/2#.
#:. I=1/2int(5x^3+3x-1)(1+cos4pix)dx#,
#=1/2int(5x^3+3x-1)dx+1/2int(5x^3+3x-1)(cos4pix)dx#,
#=1/2{5*x^4/4+3*x^2/2-x}+1/2I_1#,
#=5/8x^4+3/4x^2-1/2x+1/2I_1................(1)," where, "#
#I_1=int(5x^3+3x-1)(cos4pix)dx#,
To find #I_1#, we will repeatedly use the Rule of Integration by Parts
with #u=(5x^3+3x-1), and, v'=cos4pix#.
#:. (du)/dx=15x^2+3, and v=(sin4pix)/(4pi)#.
#:. I_1=(5x^3+3x-1)*(sin4pix)/(4pi)#
#-int(15x^2+3)*(sin4pix)/(4pi)dx#,
#=1/(4pi)(5x^3+3x-1)sin4pix-1/(4pi)I_2......(2)," where, "#
#I_2=int(15x^2+3)sin4pixdx#,
#=(15x^2+3)*{-(cos4pix)/(4pi)}-int{(30x)}{-(cos4pix)/(4pi)}dx#,
#=-3/(4pi)(5x^2+1)cos4pix+15/(2pi)I_3......(3)," where, #
#I_3=intxcos4pixdx#,
#=x*(sin4pix)/(4pi)-int{1*((sin4pix)/(4pi))dx#,
#=1/(4pi)xsin(4pix)-1/(4pi){(-cos4pix)/(4pi)}#,
#=1/(4pi)xsin(4pix)+1/(16pi^2)cos4pix............(4)#.
#(4), &, (3)rArrI_2=-3/(4pi)(5x^2+1)cos4pix#
#+15/(8pi^2)xsin4pix+15/(32pi^3)cos4pix#.
#:.," by "(2), I_1=1/(4pi)(5x^3+3x-1)sin4pix#
#-1/(4pi){-3/(4pi)(5x^2+1)cos4pix#
#+15/(8pi^2)xsin4pix+15/(32pi^3)cos4pix}#,
#=1/(4pi)(5x^3+3x-1)sin4pix+3/(16pi^2)(5x^2+1)cos4pix#
#-15/(32pi^3)xsin4pix-15/(128pi^4)cos4pix#.
Finally, from #12)#, we get,
#I=5/8x^4+3/4x^2-1/2x+1/2{1/(4pi)(5x^3+3x-1)sin4pix#
#+3/(16pi^2)(5x^2+1)cos4pix
-15/(32pi^3)xsin4pix-15/(128pi^4)cos4pix}#,
#=5/8x^4+3/4x^2-1/2x+1/(8pi)(5x^3+3x-1)sin4pix#
#+3/(32pi^2)(5x^2+1)cos4pix-15/(64pi^3)xsin4pix#
#-15/(256pi^4)cos4pix+C#.
