Evaluate the integral by making an appropriate change of variables.
$$\iint_R\left[\cos \left(\frac{y-x}{y+x}\right)\right]^2 dA$$
where $R$ is the trapezoidal region with vertices $(2,0), (3,0), (0,3)$, and $(0,2)$.
Let $u=y-x$ and $v=y+x$
The integral can rewritten as
$$\iint_R\left[\cos \left(\frac{y-x}{y+x}\right)\right]^2 d A=-\frac{1}{2} \int_2^3 \int_{-u}^u \cos ^2\left(\frac{u}{v}\right) d v d u$$
I don't think this is a good coordinate transformation, does anyone have other ideas?
On
You made a mistake in converting to $(u,v)$. We have $(x,y)=\left(\dfrac{v-u}2,\dfrac{v+u}2\right)$ so the Jacobian determinant would indeed be
$$\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix}-\frac12&\frac12\\\frac12&\frac12\end{vmatrix} = -\frac12$$
but the integration limits are mismatched. With $v=x+y$, you should actually have $2<v<3$. The other two boundaries along $x=0$ and $y=0$ correspond to
$$x=\frac{v-u}2=0 \implies u=v \\ y=\frac{v+u}2=0 \implies u=-v$$
and so your integral is
$$\begin{align*} \iint_R \cos^2\frac{y-x}{y+x} \, dA &= \frac12 \color{blue}{\int_2^3} \color{red}{\int_{-v}^v} \cos^2\frac uv \, \color{red}{du} \, \color{blue}{dv} \end{align*}$$
which is easy to compute.
Alternatively, as mentioned in an earlier comment, we can clean up the argument to $\cos$ slightly by appropriately replacing with a reciprocal. Swapping $v$ for $\dfrac1v$ transforms the integral to
$$\iint_R \cos^2\frac{y-x}{y+x} \, dA = \frac12 \int_{\tfrac13}^{\tfrac12} \int_{-\tfrac1v}^{\tfrac1v} \frac{\cos^2(uv)}{v^2} \, du \, dv$$
Ideally you'd want to further substitute $v = u/w$, but the domain of the $v$ integration includes $0$. To fix this, use the fact that the integrand is even in $v$ to simplify the integral to $\int_2^3\int_0^u\cos^2(u/v)dudv$. Then apply the substitution $v = u/w$: $$ \int_2^3\int_0^u\cos^2\frac{u}{v}dudv = \int_2^3 udu\int_1^\infty \frac{\cos^2w}{w^2}dw $$ Since the integrals are now separate, they can be evaluated individually. The first integral is easily evaluated to $5/2$. The second has no elementary form, but can be expressed in terms of the sine integral function $\operatorname{Si}$ through integration by parts and the half-angle identity $2\cos^2 \theta = 1 + \cos(2\theta)$: $$ \int_1^\infty \frac{\cos^2w}{w^2}dw = \int_2^\infty \frac{1+\cos z}{z^2}dz=-\left.\frac{1 + \cos z}{ z}\right|^\infty_2 - \int_2^\infty \frac{\sin z}{z}d z= \frac{1+\cos 2}{2} - \frac{\pi}{2}+\mathrm{Si}(2) $$ So the value of the integral is $$ \frac{5}{4}\left[2\operatorname{Si}(2)+1+\cos 2 -\pi\right]. $$