We use the Rule of Integration by Parts (IBP) in the following
form:
IBP : #intuvdx=uintvdx-int((du)/dx*intvdx)dx.#
We let, #f(x)=int(xsin2x+cos3x)dx, i.e.,#
#f(x)=intxsin2xdx+intcos3xdx,#
#=I+(sin3x)/3,# where, #I=intxsin2xdx.#
Letting #u=x and v=sin2x,# we have,
#(du)/dx=1, and, intvdx=(-cos2x)/2.#
#:. I=x(-cos2x)/2-int{1*(-cos2x)/2}dx,#
#=-1/2cos2x+1/2(sin2x)/2.#
#rArr I=-1/2cos2x+1/4sin2x.#
Altogether, #f(x)=-1/2cos2x+1/4sin2x+1/3sin3x+C.#
But, it is given that, #f(7pi/12)=14.#
#:. -1/2cos(2*7pi/12)+1/4sin(2*7pi/12)+1/3sin(3*7pi/12)+C=14.#
#:.-1/2cos(7pi/6)+1/4sin(7pi/6)+1/3sin(7pi/4)+C=14, or#
#-1/2cos(pi+pi/6)+1/4sin(p.+pi/6)+1/3sin(2pi-pi/4)+C=14, i.e.,#
#-1/2(-cos(pi/6))+1/4(-sin(pi/6))+1/3(-sin(pi/4))+C=14.#
#:. -1/2(-sqrt3/2)+1/4(-1/2)+1/3(-1/sqrt2)+C=14.#
#:. C=113/8+1/(3sqrt2)-sqrt3/4.#
#rArr f(x)=-1/2cos2x+1/4sin2x+1/3sin3x+113/8+1/(3sqrt2)-sqrt3/4.#
