Integration using change of vars

[stratusfactio]

Homework Statement

Evaluate double integral R of sin((x+y)/2)*cos((x+y)/2), where R is the triangle with vertices (0,0), (2,0), and (1,1) using change of variables u = (x+y)/2 and v = (x-y)/2

Homework Equations

Are my integrands correct? I'm getting the wrong answer...the solution is 1 - (sin 2)/2 but I'm getting 1 + (sin2)/2 - 2cos(1)sin(1)

The Attempt at a Solution

(1) Solve for x and y using the change of variables and you get y = u-v and x = u+v. Convert the (x,y) coordinates into (u,v) coordinates and for (u,v) coordinates you get new vertices (0,0), (1,1), and (1,0).

(2) SOlve for the Jacobain factor --> Factor = 2.

(3) Set up the integral: so according to the new triangle in (u,v) coordinates we see that v ranges from 0 to 1 and u ranges from v to 1. So we we have the double integral of where integral of v =(0,1), integral of u = (v,1) sin(u)*cos(v) 2dudv.

I know I have the right set up, but according to the solution, my error lies in setting up the integrals. Please help. I know this is like my 3rd question in two days =(

Homework Statement

Homework Equations

The Attempt at a Solution

 

[vela]

The integral you set up will lead to the correct answer, so you're apparently just not calculating it correctly. Show us how you evaluated the uv integral.

 

[stratusfactio]

Attempt is attached

 

[vela]

Sorry, you haven't made a mistake. You just don't recognize you have the right answer. You want to use the trig identity $\sin 2\theta = 2 \sin \theta \cos \theta$ to get
$$1+\frac{\sin 2}{2}-2\sin 1\cos 1 = 1+\frac{\sin 2}{2} - \sin 2 = 1 - \frac{\sin 2}{2}$$

 

[stratusfactio]

Thank you Vela so much!