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?

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##int_R intxcos(x+y) dR##

Where ##R = {(x, y) : 0 <= x <= pi/3, "and" 0 <= y <= pi/2}##

(I've abated the fractions.)

We can prefer whether to mix chief after a while deference to ##x## or w.r.t. ##y##. I'll meet:

##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 interior healthy, then exchange. ##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 meet

##int_0^(pi/3) [xsin(x+ pi/2) - xsin(x)] dx##

OK, perhaps that wasn't easiest. Now I accept to do 2 , but rather than re-starting, we'll preserve on this pathway.

We could restore ##sin(x+ pi/2)## by ##cos x##, but most Calculus I students won't mention that and it's probably not any simpler, so I'll preserve on after a while the collection as I accept it aloft.

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 free that when we mix ##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 healthys 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 ##


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