Finding $E[\min(X^2, 1)]$ with $X$ being standard normal I'm trying to calculate the value of $E[\min(X^2, 1)]$ with $X \sim N(0,1)$.
My attempt is that
\begin{align}
\begin{split}
E[\min(X^2, 1)] &= E[1 | X^2 \geq 1] \cdot P(X^2 \geq 1) + E[X^2 | X^2 < 1] \cdot P(X^2 < 1) \\
&= P(X^2 \geq 1) + (E[X^2] / P(X^2 < 1)) \cdot P(X^2 < 1)
\end{split}
\end{align}
but obviously this is incorrect, because it would give $P(X^2 \geq 1) + E[X^2] = P(X^2 \geq 1) + 1 > 1$, while $\min(X^2, 1) \leq 1$.
I wonder what mistake I made. I think I made a mistake when calculating $E[X^2 | X^2 < 1]$, and any help is greatly appreciated.

I prefer indicator functions over iterating the expectations, but your intuition about how to remove the $\min$ function is good.
\begin{align*}
\mathbb{E}[\min(X^2,1)] &= \mathbb{E}[\min(X^2,1) \mathbb{I}(X^2 \le 1)] + \mathbb{E}[\min(X^2,1) \mathbb{I}(X^2 > 1)] \\
&= \int_{-1}^1 x^2 \phi(x) dx  +  1 - \Phi(1) + \Phi(-1)
\end{align*}
