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$\int_{D} \frac{x-y}{1+x+y} d(x,y)$ for $D=\{(x,y)\in\Bbb R^{2}|0

I am familiar with polar coordinates and know when to use them but I struggle to structurely find other transformations to solve integrals by subsitution like this one (this one is meant to be solved like this).

Is there a general approach on how to choose the transformation or do you have to "see" it?

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    Would you be able to do it if you were given the transformation? Considering the domain $D$, a logical choice would be $u=x+y$ and $v=x-y$ since that would give (easy) fixed limits for the new variables $u$ and $v$. Coincidentally (or not ;-)), it would simplify the integrand as well. – 2017-01-12

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I'll elaborate a bit on my comment. If you have to choose a substitution, you try to make the integral easier to calculate. This can be done by either simplyfing the region of integration (and thus the limits of the variables) or by simplifying the integrand. If you're lucky, you can try doing both.

$\int_{D} \frac{x-y}{1+x+y} d(x,y)$ for $D=\{(x,y)\in\Bbb R^{2}|0

Based on the region $D$, a natural choice would be to go for a substitution of the form: $$u = x+y \; , \; v = x-y$$ This would not only simplify how the region $D$ is given as a function of the (new) variables ($0 \le u,v \le 1$), it would also simplify the integrand. Note that when performing this substitution, you shouldn't forget the Jacobian. Can you take it from there?

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    Thanks for your answer! So it does result in $2 \int_{0}^{1} \int_{0}^{1} \frac {v}{1+u }du dv$ ? – 2017-01-12
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    Indeed, well done! – 2017-01-12
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    Beautiful, thank you :) – 2017-01-12
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    I am not sure whether I messed up: Is the Jacobian $2$ or $\frac{1}{2}$? – 2017-01-12
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    @PeterGarder My apologies, I should've checked your answer more carefully; $\tfrac{\partial(u,v)}{\partial(x,y)}=2$, but you need $\tfrac{\partial(x,y)}{\partial(u,v)}=\tfrac{1}{2}$ indeed. – 2017-01-12
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You just have to "see" it, but in this case, where you are working with $x+y$ and $x-y$ all over the place, you should easily see that letting $u=x+y$ and $v=x-y$ looks promising.

So the substitution will be $$ x = \frac{u+v}{2}\\ y = \frac{u-v}{2} $$ The Jacobean of the transformation has absolute value $2$ so $$dx\,dy = 2du\,dv $$

The integral becomes $$ 2\int_{u=0}^1 \int_{=0}^1\frac{v}{1+u} du\,dv=2\int_{u=0}^1 \frac12 \frac1{1+u}du = \ln(1+1)-\ln(0+1)=\ln(2) $$

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    Thanks! But why do you choose $x = \frac{u+v}{2}, y = \frac{u-v}{2}$ instead of $x = u+v , y = u-v$? – 2017-01-12
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    Ok, now I see why. But why does $x = u+v , y = u-v$? work? Is it just coincidence/nature of the function? – 2017-01-12
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    You do substitution when it simplifies either the boundary conditions or the function being integrated. In this case, the boundary conditions are tricky, while the integral seems easy enough. So the first thing to try is to make the BC's tivial, which is what our guess does. – 2017-01-13