Calculate $\iint|cos(x+y)|dxdy$ where $0\leq x \leq π$ and $0\leq y \leq π$ using the transformation $x=u-v$ and $y=v$
This question is bothering me, as I am having problems regarding changing the region of integration, changing the limits. It would be better if full solution is provided.
Note that
$$|\cos(x+y)|=|\cos u|$$
$$dx\,dy=|J| \,du \,dv \iff dx\,dy=\begin{vmatrix} 1 & -1 \\ 0 & 1 \end{vmatrix}du\,dv\iff dx\,dy=du\,dv$$
then
$$\int_0^\pi \int_0^\pi|\cos(x+y)|\,dx\,dy =\int_0^\pi dv\int_v^{\pi+v}|\cos u|\,du= \\=2\int_0^{\frac{\pi}{2}} dv\int_v^{\frac{\pi}{2}}\cos u\,du+2\int_0^{\frac{\pi}{2}} dv\int_{\frac{\pi}{2}}^{\pi+v}-\cos u\,du= \\=2\int_0^{\frac{\pi}{2}} [\sin u]_v^{\frac{\pi}{2}}\,dv-2\int_0^{\frac{\pi}{2}} [\sin u]_{\frac{\pi}{2}}^{\pi+v} dv= \\=2\int_0^{\frac{\pi}{2}} (1-\sin v)\,dv-2\int_0^{\frac{\pi}{2}}(\sin (\pi+v)-1) dv= \\=2\int_0^{\frac{\pi}{2}} (1-\sin v)\,dv+2\int_0^{\frac{\pi}{2}}(1+\sin v) dv= \\=2[v+\cos v]_0^{\frac{\pi}{2}}+2[v-\cos v]_0^{\frac{\pi}{2}}= \\=2\left(\frac{\pi}{2}-1 \right)+2\left(\frac{\pi}{2}+1 \right)=2\pi$$