So I have this double integral...
xsin(x+y)dA
I decided to integrate in respect to dydx....
∫∫xsin(x+y)dydx
the inner integral dy goes from 0 to pi/3
the outer integral dx goes from 0 to pi/6
so I decided to integrate by parts:
u =x
du = dx
dv = sin(x+y)
v = -cos(x+y)
so I go through that process and end up getting:
-xcos(x+y)+sin(x+y)
now do I go ahead an integrate from 0 to pi/3 ?
my book has just -xcos(x+y) to begin integrating from 0 to pi/3....
so I dont know why the sin(x+y) part doesnt exist.
If it helps, I'm using the Essential Calculus: Early Transcendentals book. Section 12.1 #23