We use the Rule of Integration by Parts (IBP) in the following
form:
IBP : intuvdx=uintvdx-int((du)/dx*intvdx)dx.∫uvdx=u∫vdx−∫(dudx⋅∫vdx)dx.
We let, f(x)=int(xsin2x+cos3x)dx, i.e.,f(x)=∫(xsin2x+cos3x)dx,i.e.,
f(x)=intxsin2xdx+intcos3xdx,f(x)=∫xsin2xdx+∫cos3xdx,
=I+(sin3x)/3,=I+sin3x3, where, I=intxsin2xdx.I=∫xsin2xdx.
Letting u=x and v=sin2x,u=xandv=sin2x, we have,
(du)/dx=1, and, intvdx=(-cos2x)/2.dudx=1,and,∫vdx=−cos2x2.
:. 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.
