Interesting double integral

Question

I just had a test at calculus and it seems that nobody computed this integral. Wolfram gives me the value, so nothing, symbolab has steps currently unavailable (like always). I know the test has passed and I will never do calculus again, but still, I want to know how to compute this. $$ \iint_A\frac{2xy\sin(y)}{(1+x^4)(1+\cos^2(y))}dxdy \qquad\qquad A = [0, 1]\times[0,\pi] $$ In my thinking the first integral from left to right is the integral with respect to $y$, so the bounds will pe $0, \pi$, the second integral is with respect to $x$, in this case the bounds will be $0, 1$.

Clarification: $A = [0, 1]\cup[0,\pi]$ and let $f:A\to \Bbb R$. Compute: $$\iint_Af(x, y) dx dy $$ $A$ is just the domain of the function.

Answer 1

HINT:

For the integral $I$ with respect to $y$, enforce the substitution $y \to \pi-y$, combine results to get $I=\frac\pi{2} \int_0^\pi \frac{\sin(y)}{1+\cos^2(y)}\,dy$, and evaluate the resulting integral by making the substitution $u=\cos(y)$.

Answer 2

For the integral with respect to $x$, rather than using partial fraction decomposition, the substitution $v = x^2$, $dv = 2x \, dx$ transforms the integrand $$\frac{2x}{1+x^4} \, dx \quad \to \quad \frac{dv}{1+v^2}.$$