By inspecting the integrand, the "obvious" choice for substitution would be
u = y + x
v = y - x
Solving for x and y, we would have
x = (u - v)/2
y = (u + v)/2
in which case the Jacobian and its determinant are
[tex]J=\begin{bmatrix}x_u&x_v\\y_u&y_v\end{bmatrix}=\dfrac12\begin{bmatrix}1&-1\\1&1\end{bmatrix}\implies|\det J|=\left|\dfrac12\right|=\dfrac12[/tex]
The trapezoid R has two of its edges on the lines x + y = 8 and x + y = 9, so right away, we have 8 ≤ u ≤ 9.
Then for v, we observe that when x = 0 (the lowest edge of R), v = y ; similarly, when y = 0 (the leftmost edge of R), v = -x. So
-x ≤ v ≤ y
-(u - v)/2 ≤ v ≤ (u + v)/2
-u + v ≤ 2v ≤ u + v
-u ≤ v ≤ u
So, the integral becomes
[tex]\displaystyle\iint_R5\cos\left(7\frac{y-x}{y+x}\right)\,\mathrm dA=\int_8^9\int_{-u}^u\frac52\cos\left(\frac{7v}u\right)\,\mathrm dv\,\mathrm du[/tex]
[tex]=\displaystyle\frac52\int_8^9\frac u7(\sin7-\sin(-7))\,\mathrm du[/tex]
[tex]=\displaystyle\frac57\sin7\int_8^9u\,\mathrm du[/tex]
[tex]=\displaystyle\frac5{14}\sin7(9^2-8^2)=\boxed{\frac{85}{14}\sin7}[/tex]
