value of $ \int_0^\pi \int_0 ^\pi \vert \cos(x+y)\vert dxdy$

Question

value of $$ \int_0^\pi \int_0 ^\pi \vert \cos(x+y)\vert dxdy$$

I use the transformation: $x+y=u $ and $ y=v$, then the above integral becomes

$$ \int_{v=0}^\pi \int_{u=v} ^{\pi+v} \vert \cos(u)\vert dudv$$

after that how to do?, Please help

Answer 1

You need to calculate the integral $\int_{u=v} ^{\pi+v} \vert \cos(u)\vert du$. Just split it up in $\int_{u=v} ^{\pi} \vert \cos(u)\vert du + \int_{u=\pi} ^{\pi+v} \vert \cos(u)\vert du$. Since $|\cos(x)|$ is $\pi$-periodic, the second part will be equal to $\int_{u=0} ^{v} \vert \cos(u)\vert du$. Therefore $\int_{u=v} ^{\pi+v} \vert \cos(u)\vert du = \int_{u=0} ^{\pi} \vert \cos(u)\vert du = \int_{u=0} ^{\pi} \cos(u) du$.