Prove this multiple integrals equation let $T$ be the triangle bounded by the $x$ axis, the $y$ axis, and the line $x+y=1$.
Let $\varphi$ be a continuous function on the interval $[0,1]$, and $m,n\in \mathbb Z^+$, Show that
$$\iint_T \varphi(x+y)x^my^n\,dx\,dy=c_{mn}\int_0^1 \varphi(t)t^{m+n+1}\,dt$$
where $$c_{mn}=\int_0^1(1-t)^mt^n\,dt$$
First i tried to evaluate the right hand side integral to get the other one, but the integrand is a bit hard to evaluate, so i did this change of variables $$x=u-v\\ y=v$$
So the equation will be $$\iint_T\varphi(u)(u-v)^m v^m \,du\,dv= c_{mn}\int_0^1 \varphi(t)t^{m+n+1}\,dt $$
you may wanna ask where is the jacobian determinant? it’s equal to $1$.
Despite the change of variables, the integrand is still hard to integrate.
 A: The substitution $x=u-v,y=v$ maps the given triangle $T$ onto $\{(u,v):0\leqslant v\leqslant u\leqslant 1\}$ (well, vice versa), i.e. another triangle.
A substitution that "works" is $x=u(1-v),y=uv$ (the Jacobian is just $u$), which maps the square $\{(u,v):0<u,v<1\}$ onto the interior of $T$, and produces the claimed result after separation of variables.
A: [I was working on this for a few hours, but @metamorphy answered, but I'll still post because I don't want my work to be meaningless! So, I'll add the geometric motivation that led me to this answer at the bottom of this post]
Consider the parametrization $x+y=t$ and $y=\phi t$ where $t\in [0,1]$ and $\phi\in [0,1]$
ie, $x(t,\phi)=(1-\phi)t$ and $y(t,\phi)=\phi t$, the Jacobian det is,
$$\frac{\partial(x,y)}{\partial(t,\phi)}=\begin{vmatrix}\dfrac{\partial x}{\partial t}&\dfrac{\partial x}{\partial\phi}\\\dfrac{\partial y}{\partial t}&\dfrac{\partial y}{\partial\phi}\end{vmatrix}=\begin{vmatrix}1-\phi&-t\\\phi&t\end{vmatrix}=(1-\phi)t+\phi t=t$$
Our integral is then,
$$\begin{align}\iint\limits_T\varphi(x+y)x^my^n\,\mathrm dx\,\mathrm dy&=\int\limits_0^1\int\limits_0^1\varphi(t)(1-\phi)^mt^m\phi^nt^nt\,\mathrm dt\,\mathrm d\phi\\&=\left(\int_0^1 \phi^m(1-\phi)^n\,\mathrm d\phi\right)\left(\int_0^1\varphi(t)t^{m+n+1}\,\mathrm dt\right)\\&=c_{mn}\int_0^1\varphi(t)t^{m+n+1}\,\mathrm dt\end{align}$$ as desired.

Now, for the motivation:
Think of the integration region $T=\{(x,y)\mid x+y\le 1,~x\ge 0,~y\ge 0\}$ as a union of straight lines: $$T=\bigcup_{t\in [0,1]}\{(x,y)\mid x+y=t,~x\ge 0,~y\ge 0\}$$
Each point on such a line $x+y=t$ can be uniquely represented by the perpendicular distance from the origin to the line (which is $t/\sqrt 2$) and the projection of that point on y-axis, which is simply $y$ (or equivalently, the projection on x-axis), and as such, every point of $T$ has a unique representation in terms of the perpendicular distance $t/\sqrt 2$ and the projection onto one of the axes. The projection $y$ will range from $0$ to $t$, so we can introduce another parameter $\phi$ as a scaling parameter with values in $[0,1]$ so that each point $(x,y)$ (Cartesian) of $T$ can be uniquely mapped to coordinates $(\sqrt 2t,\phi t)$ where $t$ ranges from $0$ to $1/\sqrt 2$, or equivalently to co-ords $(t,\phi t)$ where $t$ ranges from $0$ to $1$
So, we now have a one-to-one map between $T$ and the unit square (as @metamorphy wrote so elegantly; I instead drew figures and scratched my head for half an hour to write $x,y$ in terms of $t,\phi$ while trying to reverse-engineer the process, till I finally worked it out :') ) and the rest follows as above.
[will be adding the diagrams I made while working on this]
