Solving double integral $\iint_A \frac{\tan(x+y)}{x+y}dxdy$

Question

I have to solve this double integral $$\iint_A \frac{\tan(x+y)}{x+y}dxdy$$ where $A=\{(x,y)\in\mathbb{R}^2| x+y\le 1;y>0;x>0\}$.

I tried with a linear substitution given by: $$\cases{x+y=t\\x-y=u}\Rightarrow\cases{x=\frac{t+u}{2}\\y=\frac{t-u}2}$$

but then I am not able to solve the integral nor to find the extremes.

Do you know some other substitution or some new way to manipulate the integrand?

Answer 1

Your substitution is going to work well if you find out the limits.

To limit $t$ is easy: $t=x+y\le 1$, and $x,y>0$ gives $t=x+y>0$. Hence, $0<t\le 1$.

To limit $u$: use that $2x=t+u>0$ and $2y=t-u>0$. What are bounds for $u$?

$-t<u<t$.