$ \iint_{R} \cos((3-y)^2) dx dy$ over the region $R=\{(x,y): |x-1|+x \le y \le x+1)\}$

Question

Not sure how to start this question, since $\cos((3-y)^2)$ cannot be easily integrated. I believe there is a change of variable that needs to take place but I'm not sure what values to use.

Answer 1

Note that you need to calculate $$\iint_{\mathscr{R}:\{(x,y):|x-1|-x\leq y\leq x+1\}}\cos((3-y)^{2})dxdy$$ Now, by definition we need to evaluate $$\int_{a}^{b}\int_{g_{1}(y)}^{g_{2}(y)}\cos((3-y)^{2})dxdy$$ Since that region of integration is $$\mathscr{R}:\{(x,y):|x-1|-x\leq y\leq x+1\}$$ so, we get $$g_{1}(y)=y-1 \quad \text{and} \quad g_{2}(y)=\frac{y+1}{2}$$ and $$a=1 \quad b=3$$ Therefore, \begin{eqnarray*} \iint_{\mathscr{R}:\{(x,y):|x-1|-x\leq y\leq x+1\}}\cos((3-y)^{2})dxdy&=&\int_{1}^{3}\int_{y-1}^{\frac{y+1}{2}}\cos((3-y)^{2})dxdy\\ &=&\int_{1}^{3}\cos((3-y)^{2})\left[\frac{y+1}{2}-(y-1) \right]dy\\ &=&\int_{1}^{3}\cos((3-y)^{2})\frac{y+1}{2}dy-\int_{1}^{3}\cos((3-y)^{2})(y-1)dy\\ &=&\left(\sqrt{2\pi}C\left[2\sqrt{\frac{2}{\pi}} \right]-\frac{\sin(4)}{4}\right)-\left(\sqrt{2\pi}C\left[2\sqrt{\frac{2}{\pi}} \right]-\frac{\sin(4)}{2}\right)\\ &=&\boxed{\frac{\sin(4)}{2}-\frac{\sin(4)}{4}} \end{eqnarray*} where $C(x)$ is theFresnel $C$ integral.