How do you calculate the double integral of ##(xcos(x+y))dr## where r is the region: 0 less than or equal to x less then or equal to ##(2pi)/6##, 0 less than or equal to y less than or equal to ##(2pi)/4##?

##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 ##