Calculate $\int^{1/2}_0\int^{1-x}_x (x+y)^9(x-y)^9 \, dy \, dx$ How can I find the following integral:
$$\int^{1/2}_0 \int^{1-x}_x (x+y)^9(x-y)^9 \, dy \, dx $$
My thoughts:
Can we possibly convert this to spherical or use change of variables? 
 A: according to the shape of the area of integration and the shape of the function that is under integral, the easiest answer is to define variables $u=x+y$ and $v=x-y$then we have:
$$\frac{1}{J}=\frac{\partial(u,v)}{\partial(x,y)}=\begin{vmatrix}1&1\\1&-1\end{vmatrix}\Rightarrow |J|=\frac{1}{2}$$
and the borders of the area of integration are $u=-v,u=1,v=0$ just draw the shape of the problem to understand this part.then:
$$\int^{1/2}_0 \int^{1-x}_x (x+y)^9(x-y)^9 \, dy \, dx=\int^{0}_{-1}\int^1_{-v} \frac{1}{2}u^9v^9\,du\,dv=\frac{-1}{400}$$

Edit: to answer the question asked in the comments:
why did I use $\frac{1}{|J|}$?
consider the analog situation for a 1-D integral where we have to use change of variables method:
$$\int\frac{x\,dx}{\sqrt{1-x^4}}$$
we have $u=x^2\Rightarrow du=2x\,dx$
meaning that we have $u$ as a function of $x$ so we can calculate $du=\frac{du}{dx}dx$ but in order to substitute $dx$ in the original integral we write $du=2x\,dx\Rightarrow dx=\frac{1}{2x}du$ which means that we need $dx=\frac{dx}{du}du$ to substitute in the original integral so
In a 1-D integral we have $u$ as a function of $x$ so we can calculate $\frac{du}{dx}$ but to substitute in the original integral we need $\frac{dx}{du}$
here is the same situation:
to substitute in the original integral, we need:
$$dx\,dy=\frac{\partial(x,y)}{\partial(u,v)}du\,dv=|J|du\,dv$$
But most of the time we have $u$ and $v$ as a function of $x$ and $y$:
$$u=f(x,y),v=g(x,y)$$
so we can compute 
$$\frac{1}{|J|}=\frac{\partial(u,v)}{\partial(x,y)}$$
A: Hint: $$(x+y)^9(x-y)^9=((x+y)(x-y))^9=(x^2-y^2)^9$$
A: How about a change of variable like $u=x+y, v=x-y$?
The Jacobian is $-\frac 12$, and the area of integration is the triangle bounded by the lines $x=y, x+y=1, x=0$ 
This translates as: $v$ varies from $v=0$ to $ v=u$ for the inner integral, and $u=0$ to $u=1$ for the outer integral. 
Therefore we evaluate $$\int_{u=0}^{u=1}\int_{v=0}^{v=u}u^9v^9(-\frac 12)dvdu$$
The final answer is $-\frac{1}{400}$ 
A: \begin{align}
u & = x+y \\
v & = x-y
\end{align}
$$
du\,dv = \left|\frac{\partial (u,v)}{\partial(x,y)}\right|\,dx\,dy = 2\,dx\,dy
$$
\begin{align}
& \int^{1/2}_0 \int^{1-x}_x (x+y)^9(x-y)^9 \, dy \, dx \\[10pt]
= {} & \int_0^1 \left( \int_{u-1}^0  u^9 v^9 2\,dv \right) \,du \\[10pt]
= {} & \int_0^1 u^9 \frac{(-(u-1)^{10})} 5  \, du 
\end{align}
