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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.

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    Do you mean $A = [0,1]\times[0, \pi]$? – 2017-01-27
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    the integral over $x,y$ separete – 2017-01-27
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    The first step is to split the integrand. If the integrand is of the form $f(x)\cdot g(y)$, the double integral is equal to the product of the riemann integrals over $f(x)$ and $g(y)$ – 2017-01-27
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    for the integral over $y$ notice that $d\cos(y)=\sin(y) dy$ – 2017-01-27
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    @tired First exploit the symmetry to get rid of that pesky $y$ as per my posted hint – 2017-01-27

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

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    (+1) our combined hints ought to do it. And Happy New Year. -Mark – 2017-01-27
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    I've did that. And you're left with something hideous. I'm not sure that's the right mindset – 2017-01-27
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    @vladpotra The integral $\int \frac1{1+v^2}\,dv =\arctan(v)+C=\arctan(x^2)+C$ is a "look up" integral. It will be useful in evaluating the integral over $y$ after applying my hints. – 2017-01-27
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    @Dr.MV I know it helps, that's also the way I think. But still, we get to something really strange after that. – 2017-01-27
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    @vladpotra Evaluate $\arctan(x^2)$ at the upper limit $1$ and the lower limit $0$. Taking the difference gives $\pi/4$. The integral over $y$ will result in something proportional to $\pi^2$. – 2017-01-27
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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 $u=\cos(y)$.