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