An obvious substitution would be to take u = x - y and v = x + 2y, so that 0 < u < 1 (which follows from y = x ==> x - y = 0 and y = x - 1 ==> x - y = 1), and 0 < v < 1.
Solve for x and y in terms of u and v :
u - v = (x - y) - (x + 2y) = -3y ==> y = (v - u)/3
2u + v = 2 (x - y) + (x + 2y) = 3x ==> x = (2u + v)/3
Compute the Jacobian and its determinant:
[tex]J=\begin{bmatrix}D_ux&D_vx\\D_uy&D_vy\end{bmatrix}=\begin{bmatrix}\dfrac23&\dfrac13\\\\-\dfrac13&\dfrac13\end{bmatrix} \implies |\det(J)|=\dfrac13[/tex]
Now in the integral, we have
[tex]\displaystyle\iint_R\frac{x+2y}{\cos(x-y)}\,\mathrm dx\,\mathrm dy = \frac13\int_0^1\int_0^1\frac{v}{\cos(u)}\,\mathrm du\,\mathrm dv \\\\ \displaystyle= \frac13\left(\int_0^1v\,\mathrm dv\right)\left(\int_0^1\sec(u)\,\mathrm du\right) \\\\ \displaystyle= \frac13\bigg(\frac{v^2}2\bigg)\bigg|_0^1\bigg(\ln|\sec(u)+\tan(u)|\bigg)\bigg|_0^1 \\\\ \displaystyle= \boxed{\frac16\ln(\sec(1)+\tan(1))}[/tex]
