# Thread: Double integration with substitution

1. ## Double integration with substitution

Use substitution to find $\displaystyle \int \int \cos (\frac{x-y}{x+y})dx dy$ over region D, where D is the triangle with vertices (0,0), (0,1), (1,0).

I think the required integral would then be the following, but have no clue as to what substitution is required.

$\displaystyle \int_0^1 \int_0^{1-x} \cos (\frac{x-y}{x+y})dy dx$

I would like someone to check that my limits are correct and then point me in the right direction for a substitution, explaining how I should have come up with it. Thanks!

2. ## Change of variables

The key is to use the change of variables theorem, such as when one changes between Cartesian and other (such as polar) coordinate systems.

Let us define new coordinates x',y' whose axes are rotated 45 degrees clockwise from the x,y axes. Then $\displaystyle x'=\frac{x-y}{\sqrt{2}}$ and $\displaystyle y'=\frac{x+y}{\sqrt{2}}$. We can see that the Jacobian determinant $\displaystyle \left|\frac{\partial(x,y)}{\partial(x',y')}\right| =1$, so dA=dxdy=dx'dy'. Note that in the x',y' coordinates, the region is the triangle with vertices (0,0), ($\displaystyle -\frac{1}{\sqrt{2}}$,$\displaystyle \frac{1}{\sqrt{2}}$), and ($\displaystyle \frac{1}{\sqrt{2}}$,$\displaystyle \frac{1}{\sqrt{2}}$)

Making the substitution, we see $\displaystyle \cos \left(\frac{x-y}{x+y}\right)=\cos \left(\frac{\sqrt{2}x'}{\sqrt{2}y'}\right)=\cos \left(\frac{x'}{y'}\right)$
Thus we have
$\displaystyle \int \int \cos \left( \frac{x-y}{x+y} \right) \,dA=\int_{0}^{\frac{1}{\sqrt{2}}} \int_{-y'}^{y'} \cos \left(\frac{x'}{y'}\right)\,dx'\,dy'$

You should be able to do that integral.

--Kevin C.