Answer
$\dfrac{1-\cos (1)}{2}$
Work Step by Step
The region $D$ can be defined as:
$D=\left\{ (x, y) | 0 \leq y \leq x^2, \ 0 \leq x \leq 1 \right\}
$
$\iint_{D} x \cos y dA=\int_{0}^{1} \int_{0}^{x^2} x \cos y \ dy \ dx \\= \int_0^1 x \sin x^2 d x $
Consider $a=x^2$ and $ da=2x dx$
$$\iint_{D} x \cos y dA=\int_0^1 \sin a \dfrac{da}{2} \\ =\dfrac{1}{2} \int_{0}{1} \sin a da \\ = \dfrac{1} {2} \times [ -\cos a]_0^1 \\= \dfrac{1-\cos (1)}{2}$$
