How to find $I=\iint_{-1\le x\le 1,x\le y\le 1}y\left(e^{-|y|^3}+\sin{x}\right) \, dx \, dy$ 
Find $$I=\iint_{-1\le x\le 1,x\le y\le 1}y\left(e^{-|y|^3}+\sin{x}\right) \, dx \, dy.$$

I think this integral might be evaluated using $\displaystyle \int_{-a}^{a}f(x) \, dx=0$  if $f(x)=-f(-x)$.
Does anyone have any nice methods?
 A: Hint: The only thing which can help us here is to change the limits of the integrals. I mean $$\int_{-1}^{1}\int_x^1f(x,y) \, dy \, dx\longrightarrow\int_{-1}^{1}\int_{-1}^yf(x,y) \, dx \, dy$$

A: Let 
$$B:=\{(x,y)\>|\>-1\leq x\leq 1,\ x\leq y\leq 1\}$$ and put $$B_x:=\{y\>| \>(x,y)\in B\}=[x,1]\qquad(-1\leq x\leq 1)\ ,$$
$$B^y:=\{x\>|\>(x,y)\in B\}=[-1,y]\qquad(-1\leq y\leq 1)\ .$$
Then
$$\int\nolimits_B y\>\sin x\ {\rm d}(x,y)=\int_{-1}^1 \left(\sin x\int\nolimits_{B_x} y\ dy\right)\ dx={1\over2}\int_{-1}^1 \sin x\>(1-x^2)\ dx=0\ .$$
It follows that your integral ($=:J$) is equal to (this time we integrate first with respect to $x$):
$$\int\nolimits_B y\>e^{-|y|^3}\ {\rm d}(x,y)=\int_{-1}^1 \left( y\>e^{-|y|^3} \int\nolimits_{B^y} dx\right)\ dy=\int_{-1}^1 y\>e^{-|y|^3}(y+1)\ dy\ .$$
By symmetry, we finally obtain
$$J=2\int_0^1 y^2 e^{-y^3}\ dy=-{2\over3}e^{-y^3}\biggr|_0^1={2\over3}(1-e^{-1})\ .$$
A: 
Following method is relative simple due to the exploitation of the symmetry. 
Divide the region into two as above figure. Blue region is the upper triangle:
$$
\{ 0\leq y\leq 1,-y\le x\le y \}.
$$
Cyan region is the left triangle:
$$
\{ -1\leq x\leq 0,x\le y\le -x \}.
$$
The integration in the blue region is:
$$I_1 = \iint_{-1\le x\le 1,-y\le x\le y\le 1}y\left(e^{-|y|^3}+\sin{x}\right) \, dx \, dy
\\
 = \int_{0}^1\int_{-y}^y\left( y e^{-y^3}+ \color{\red}y\color{\red}{\sin{x}}\right) \, dx \, dy.$$
Red term vanishes because $\sin x$ is odd:
$$
\int_{-y}^y y\sin{x} \,dx = 0.
$$
Hence
$$I_1 =\int_{0}^1\int_{-y}^y y e^{-y^3}\, dx \, dy = \int_{0}^1 2y^2 e^{-y^3} \,dy= \frac{2}{3}(1-e^{-1}).$$
The integration in the cyan region is:
$$
I_2 = \iint_{-1\le x\le 0,x\le y\le -x}y\left(e^{-|y|^3}+\sin{x}\right) \, dx \, dy\\
= \int_{0}^{-1}\int_{-x}^x\left( \color{\red}{y} e^{-\color{\red}{|y|}^3}+ \color{\red}{y}\sin{x}\right) \, dy \, dx.
$$
Both terms vanishes, because $ y e^{-|y|^3}$ and $y\sin{x}$ are odd with respect to $y$:
$$
\int_{-x}^x\left( y e^{-|y|^3}+ y\sin{x}\right) \, dy = 0.
$$
Hence:
$$
I = I_1 + I_2 = \frac{2}{3}(1-e^{-1}).
$$
