Answer
$\dfrac{2}{3} \sin 1$
Work Step by Step
We can define the domain $D$ in the Type-II using horizontal cross-sections as follows:
$D=\left\{ (x, y) | y/2 \leq x \leq 1, \ 0 \leq y \leq 2 \right\}
$
Therefore, $\iint_{D} f(x,y) dA=\int_{0}^{2} \int_{1}^{y/2} y \cos (x^3-1) \ dx \ dy$
We can define the domain $D$ in the Type-1 using vertical cross-sections as follows: $D=\left\{ (x, y) | 0 \leq y \leq 2x, \ 0 \leq x \leq 1 \right\}
$
Therefore, $\iint_{D} f(x,y) dA=\int_{0}^{1} \int_{0}^{2x} y \cos (x^3-1) dy \ dx \\= \int_0^1 [\dfrac{y^2}{2} \cos (x^3-1)]_0^{2x} \ dx \\= \int_0^1 2x^2 \cos (x^3-1) \ dx $
Let us consider $a=x^3 -1 \implies da=3x^2 dx$
Now, $\int_0^1 2x^2 \cos (x^3-1) \ dx=\int_{-1}^{0} 2 \cos a \dfrac{da}{3} \\= [\dfrac{2}{3} \sin a ]_{-1}^0 \\=\dfrac{2}{3} \sin (0) -\dfrac{2}{3} \sin (-1) \\=\dfrac{2}{3} \sin 1$
