Evaluate $$I=\iint\limits_R \sin \left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)\, dA,$$ where $R$ is the triangle with vertices $(0,0),(2,0)$ and $(1,1)$.
Hint: use $u=\dfrac{x+y}{2},v=\dfrac{x-y}{2}$.
Can anyone help me with this question I am very lost. Please help
I know you can make the intergal $\sin(u)\cos(v)$, but then what to do?
Best Answer
There is another way to do this. You might notice that the integrand is
$$2 \sin{\left(\frac{x+y}{2}\right)} \cos{\left(\frac{x-y}{2}\right)} = \sin{x} + \sin{y}$$
You may then integrate this over the triangle directly:
$$\frac12 \int_0^1 dx \, \int_0^x dy \, [\sin{x} + \sin{y}] + \frac12 \int_1^2 dx \, \int_0^{2-x} dy \, [\sin{x} + \sin{y}] $$
Note that I formed the integration boundaries from the equations of the lines formed from the vertices of the triangle. Note also that I had to break this in two: one for the left side and one for the right.
You may then evaluate this in terms of single integrals by integrating over $y$; I get
$$\frac12\int_0^1 dx \, [x \, \sin{x} + 1 - \cos{x}] + \frac12\int_1^2 dx \, [(2-x) \sin{x} + 1-\cos{(2-x)}]$$
I will let you take it from here.