What is f(x) = int xsin2x+cos3x dx if f((7pi)/12)=14 ?

1 Answer
Dec 13, 2017

-1/2cos2x+1/4sin2x+1/3sin3x+113/8+1/(3sqrt2)-sqrt3/4.

Explanation:

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.