Let $R\subset S$ be an integral extension of unital commutative rings, and let $I$ be an ideal in $R$. I suppose that if $I$ is prime, the following can be deduced from Lying over and Going up, but does it hold in general?
$R\cap IS\subset I$.
(Obviously, $\supset$ holds.)
To prove, I would primitively start by assuming there exists a $x\in IS\cap R\setminus I$, then $x=s_1i_1+...+s_ni_n$, bump. To find a counterexample, I have no idea.
As @Youngsu pointed out, for a counterexample one can use $R=k[x^2,x^3]\subset k[x]=S,$ with $I=(x^2)_R$. This serves as a counterexample since $(x^2)_R \not\ni x^3\in (I)_S\cap R$ and $x$ is integral over $R$ with equation $R[X]\ni p(X)=X^2-x^2$. Note that one can use any nonzero ring instead of $k$, or also the formal power series instead of the polynomial ring.
For $I=P$ prime I suppose this proof: From lying over and going up, there is a prime lying over P, which clearly must contain $(P)_S$ and the claim follows.
Note: I used different notations, $(P)_S=PS$.