##int_R intxcos(x+y) dR##
Where ##R = {(x, y) : 0 <= x <= pi/3, “and” 0 <= y <= pi/2}##
(I’ve reduced the fractions.)
We can choose whether to integrate first with respect to ##x## or w.r.t. ##y##. I’ll find:
##int_0^(pi/3) int_0^(pi/2) xcos(x+y) dy dx = int_0^(pi/3) [int_0^(pi/2) xcos(x+y) dy] dx##
I’ll do the inner integral, then substitute. ##int_0^(pi/2) xcos(x+y) dy = xsin(x+y)]_(y=0)^(y = pi/2) = xsin(x+ pi/2) – xsin(x)##
Now we need to find
##int_0^(pi/3) [xsin(x+ pi/2) – xsin(x)] dx##
OK, maybe that wasn’t easiest. Now I have to do 2 , but rather than re-starting, we’ll keep on this path.
We could replace ##sin(x+ pi/2)## by ##cos x##, but most Calculus I students won’t notice that and it’s probably not any simpler, so I’ll keep on with the problem as I have it above.
First Integral:
##int_0^(pi/3) x sin(x+ pi/2) dx##
Let ##u=x## and ##dv = sin(x+ pi/2) dx##
so ##du = dx## and ##v = -cos(x+ pi/2)##
##int_0^(pi/3) x sin(x+ pi/2) dx = -xcos(x + pi/2) + int cos(x + pi/2) dx##
## = -xcos(x + pi/2) + sin(x + pi/2)]_0^(pi/3)##
##= [- (pi)/3 cos((5 pi)/6) + sin((5 pi)/6)] – [0 + sin(pi/2)]##
##= – pi/3 (- sqrt3 / 2) +1/2 – 1 = (pi sqrt3)/6 -1/2 = (pi sqrt 3 -3)/6##
Second Integral:
It should be clear that when we integrate ##int_0^(pi/3) [ – xsin(x)] dx## we’ll get
## = – (-xcos(x) + sin(x))]_0^(pi/3) = = xcos(x) – sin(x))]_0^(pi/3)##
Which evaluates to:
##(pi)/3 1 /2 – sqrt3/2 -[0cos (0) – sin(0)] = (pi – 3sqrt3)/6##
Adding the two integrals gives us:
##int_0^(pi/3) [xsin(x+ pi/2) – xsin(x)] dx = (pi sqrt 3 -3)/6+(pi – 3sqrt3)/6 = (pi sqrt3 – 3 sqrt3 + pi -3)/6 ##
