Let $D$ be the region bounded by $x+y = 1$, $x= 0$, $y = 0$. Use the result of Exercise 19 to show that $$\iint_D\cos\bigg(\frac{x-y}{x+y}\bigg)~dx~dy = \frac{\sin 1}{2} $$ and graph $D$ on an $xy$ plane and a $uv$ plane, with $u = x-y$ and $v = x+y$.
I got the Jacobian determinant is $\dfrac{1}{2}$, but how do I get the limits of integration for $u$ and $v$?
On
Let $D$ be the region bounded by $x+y = 1$, $x= 0$, $y = 0$. with $u = x-y$ and $v = x+y$.
$D$ is $\mathop\triangle\limits_{x,y}(0,0)(0,1)(1,0)$, and the transformation is a rotation and scale shift to $D^*$ which is $\mathop\triangle\limits_{u,v}(0{-}0,0{+}0)(0{-}1,0{+}1)(1{-}0,1{+}0)$.
Alternatively:
As $x=(u+v)/2, y=(v-u)/2$ , then $~D=\{(x,y): x\geq 0, y\geq 0, x+y\leq 1\}\\D^\ast{=\{(u,v): u{+}v\geq 0, v{-}u\geq 0, (u{+}v){+}(v{-}u)\leq 2\} \\= \{(u,v): {0\leq v\leq 1, -v\leq u\leq v}\}}$
For convenience, let's write your change of variable as: $$x=\frac{u+v}{2},\qquad y=\frac{v-u}{2} \tag{1}$$
The most obvious thing to do when trying to find the region $D^*$ in the $uv$-plane corresponding to region $D$ is to sketch the region of integration:
By substitution of equations $(1)$, we notice that the image lines of $y=0$, $x=0$ and $y=1-x$ in the $uv$-plane is $v=u$, $v=-u$ and $v=1$ respectively, as shown below: $$y=0 \implies \frac{v-u}{2}=0 \implies u=v$$ $$x=0 \implies \frac{u+v}{2}=0 \implies u=-v$$ $$y=1-x \implies \frac{v-u}{2}=1-\frac{u+v}{2}\implies v-u=2-u-v \implies v=1$$
Let's now sketch the new region $D^*$:
Notice that it is more convenient to describe $D^*$ as a type II plane region: $$D^*=\{(u,v)\mid 0\leq v \leq 1, -v\leq u \leq v\}$$ Therefore, the integral becomes: $$\iint_D \cos\left(\frac{x-y}{x+y}\right)~dA=\iint_{D^*} \cos(u/v)\cdot \left|\frac{\partial(x,y)}{\partial(u,v)}\right|~du~dv=\int_0^1 \int_{-v}^v \frac{\cos(u/v)}{2}~du~dv=\frac{\sin(1)}{2}$$ The last line is easily verifiable, and is left as an exercise for the reader.