how to calculate $\int\int \frac{\sin(x)-\sin(y)}{xy+1} dx~dy$
bounded by $x^2 = y ; ~y^2=x$.
i dont know how to change the variables such that the integral is simple.
maybe $\sin(x) - \sin(y) = 2\cos(\frac{x+y}{2})\sin(\frac{x-y}{2})$ would help ?
its not just about changing the variable , the $~~ (\sin~ \cos)~~$ making it difficult
any ideas \ hints ?
Given the function $f : D \to \mathbb{R}$ of law:
$$ f(x,\,y) := \frac{\sin x - \sin y}{1 + x\,y} $$
where is it:
$$ D := \left\{ (x,\,y) \in \mathbb{R}^2 : 0 \le x \le 1, \; x^2 \le y \le \sqrt{x} \right\}, $$
since $D$ enjoys symmetry $\mathcal{S}(x,\,y) = (y,\,x)$ and the function $f$ is odd with respect to $S$, so:
$$ \iint_D f = 0\,. $$