A substitution that will simplify this integrand

Question

$$\iint_{R}^{} x \sin(6x + 7y) - 3y \sin(6x + 7y) dA$$

So I chose $u = 3y$ and $v = 6x + 7y$. So then $x$ will be replaced with $\frac{3u - 7v}{18}$. It seems correct, but does this truly "simplify" the integrand? Or have I done something incorrectly?

Answer 1

Note that after factoring, we have

$$\iint_{R}^{} x \sin(6x + 7y) - 3y \sin(6x + 7y)\,dA = \iint_{R}^{} (x-3y)\,\sin(6x + 7y)\,dA.$$

So a better substitution would probably be:

$$\begin{cases} u=x-3y,\\v=6x+7y.\end{cases}.$$

Answer 2

Hint: Choose $u=6x+7y$ and $v=x-3y$ to get

$$\int\int vsin(u) (\frac{1}{25})dudv$$ where $\dfrac{1}{25}=\dfrac{\partial(x,y)}{\partial(u,v)}$