Cumulative incidence of X Suppose the joint survival function of the latent failure times for two competing risks, $X$ and $Y$, is $S(x,y)=(1-x)(1-y)(1+0.5xy)$, $0<x<1$, $0<y<1$. Find the cumulative incidence function of $X$?
I first solved the marginal cumulative distribution function of $X$: $(1-x)$. Then I tried to find the joint density function: $1.5-x-y+2xy$, but I am unable to determine how to properly integrate this to find the cumulative incidence function of $X$.

For $0 \leq x \leq 1$ the cumulative incidence of $X$ is defined as
$$
\mathbb P \left (X \leq x, X \leq Y \right)
$$
To compute this probability we need to integrate the joint density of $(X,Y)$,
$$
f(x,y) = \frac{3}{2} -x-y+2xy
$$
over the set $\mathcal{A} \equiv \{(u,v) \in [0,1]^2 \mid u \leq x \wedge u \leq v \} $
That is,
\begin{align*}
\mathbb P \left (X \leq x , X \leq Y \right) &= \int_\mathcal{A} f(u,v)\text{d}u\text{d}v \\
&= \int_0^x \left( \int_u^1 f(u,v)\text{d}v \right )\text{d}u \\
&= \int_0^x \left( \int_u^1  \left(\frac{3}{2} -u-v+2uv \right)\text{d}v \right )\text{d}u \\
&= \int_0^x \left( \frac{3}{2}(1-u) -u+u^2 - \left[ \frac{v^2}{2} \right]_u^1 +2u \left[ \frac{v^2}{2} \right]_u^1 \right) \text{d}u\\
&=\int_0^x \left( \frac{3}{2} - \frac{3}{2} u -u+u^2 - \left( \frac{1}{2}-\frac{u^2}{2}  \right) +2u \left( \frac{1}{2} - \frac{u^2}{2} \right) \right) \text{d}u\\
&=\int_0^x \left( 1- \frac{3u}{2} + \frac{3u^2}{2} -u^3\right) \text{d}u\\
&= \left[ u - \frac{3u^2}{4} + \frac{u^3}{3} - \frac{u^4}{4} \right ]_0^x \\
&=  x - \frac{3x^2}{4} + \frac{x^3}{3} - \frac{x^4}{4} \\
&= \frac{12x -9x^2 + 6x^3 - 3x^4}{12}.
\end{align*}
A quick check, take  $x=1$ we have
$$
\frac{12x -9x^2 + 6x^3 -3x^4}{12} = \frac{1}{2}
$$ and
$$
\mathbb P \left (X \leq 1, X \leq Y \right) \stackrel{(1)}{=} \mathbb P(X\leq Y) \stackrel{(2)}{=}\frac{1}{2}
$$
Here $(1)$ comes from the fact that since $X \leq 1$ with probability $1$, the event $\{X \leq 1 \cap X \leq Y\}$ has the same probability as the event $\{X \leq Y\}$ and since $X$ and $Y$ are somehow "symmetric" we have $\mathbb P(X\leq Y) = \mathbb P(Y\leq X) = \frac{1}{2}$ hence equality $(2)$.
