Suppose that, #I=int(x^2+5)cos2xdx.#
We will use the following Rule of Integration by Parts (IBP) :
#" IBP ": intu*vdx=uintvdx-int{(du)/dxintvdx}dx.#
We take, #u=x^2+5, and, v=cos2x.#
#:. (du)/dx=2x, and, intvdx=(sin2x)/2.#
#:. I=(x^2+5)/2*sin2x-int{(2x)(sin2x)/2}dx,#
#=1/2(x^2+5)sin2x-J,"... ...(1), where, "J=intxsin2xdx.#
To evaluate #J,# we reuse IBP, this time, with,
#u=x, v=sin2x rArr (du)/dx=1, intvdx=-1/2*cos2x.#
# :. J=-x/2*cos2x-int{(1)(-1/2*cos2x)}dx,#
#=-x/2*cos2x+1/2intcos2xdx,#
#=-x/2*cos2x+1/2*1/2sin2x,#
#:. J=-x/2*cos2x+1/4*sin2x.........(2).#
From #(1), and, (2),# we have,
#I=1/2(x^2+5)sin2x+x/2*cos2x-1/4*sin2x+C, or, #
#I=1/4(2x^2+9)sin2x+x/2*cos2x+C.#
Enjoy Maths.!
