If X and Y are two independent standard normal r.v.s, calculate $P(X+Y \in [0,1] \mid X \in [0,1])$ I'm not sure how to solve this. This is my attempt so far.
So if X,Y two independent standard normal r.v.s, we have:
$$\mathbb{P}(X+Y\in [0,1] \mid X \in [0,1])=\frac{\mathbb{P}(\{X+Y\in [0,1]\} \cap \{X \in [0,1]\})}{\mathbb{P}(X \in [0,1])}.$$
Moreover, we have:
\begin{split} \mathbb{P}(\{X+Y\in [0,1]\} \cap \{X \in [0,1]\}) = {} &  \int_0^{1}dx\frac{1}{\sqrt{2 \pi}}e^{-\frac{x^2}{2}}\int_{-x}^{1-x} dy\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}.
\end{split}
To calculate the integral, it looks like it might be better to switch to polar coordinates (?). Then we have:
\begin{split} \int_0^{1}dx\frac{1}{\sqrt{2 \pi}}e^{-\frac{x^2}2}\int_{-x}^{1-x} dy \frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}  = & \frac{1}{2\pi}\int_{-\frac\pi4}^{0} \, d \varphi \int_0^{\frac{1} {  \cos \varphi}}r e^{-\frac{r^2}2} \, dr \\
& +  \frac{1}{2\pi}\int_{0}^{\frac\pi2} \, d \varphi \int_0^{\frac1 { \sin \varphi + \cos \varphi}}r e^{-\frac{r^2}2} \, dr \\
= &- \frac{1}{2\pi}\int_{-\frac\pi4}^{0} \, d \varphi \int_0^{-\frac1 { 2\cos^2 \varphi}} e^t \, dt \\
&  - \frac{1}{2\pi}\int_{0}^{\frac{\pi}{2}}d \varphi \int_{0}^{-\frac{1}{2 ( \sin \varphi + \cos \varphi) ^2}} e^t\,dt, \\
\end{split}
and I don't know what to do from here. Although, I'm not even sure if my procedure doesn't have any mistakes. Thanks for any insights.

Write
$$ \{ X +Y \in [0, 1], X \in [0, 1]\} = \{ (X, Y) \in D_1 \} \cup \{ (X, Y) \in D_2 \}, $$
where
\begin{align*}
D_1 &= \{(x, y) : 0 \leq x \leq 1 \text{ and } 0 \leq y \leq 1 \text{ and } 0 \leq x+y \leq 1 \} \\
D_2 &= \{(x, y) : 0 \leq x \leq 1 \text{ and } -1 \leq y \leq 0 \text{ and } 0 \leq x+y \leq 1 \}.
\end{align*}
$\hspace{14.5em}$
Then by symmetry,
\begin{align*}
\Bbb{P}(X +Y \in [0, 1], X \in [0, 1])
&= \Bbb{P}((X, Y) \in D_1) + \Bbb{P}((X, Y) \in D_2) \\
&= \tfrac{1}{4}\Bbb{P}((X, Y) \in \tilde{D}_1) + \tfrac{1}{8}\Bbb{P}((X, Y) \in [-1,1]^2)
\end{align*}
where
$$ \tilde{D}_1 = \{(x, y) : -1 \leq x + y \leq 1 \text{ and } -1 \leq x - y \leq 1 \} $$
$\hspace{11em}$
is the square with corners $(\pm 1, 0)$ and $(0, \pm1)$. Finally, using the fact that the law of $(X, Y)$ is rotation invariant, we can replace $\tilde{D}_1$ by its $45^{\circ}$ rotation without affecting the probability:
\begin{align*}
\Bbb{P}(X +Y \in [0, 1], X \in [0, 1])
&= \tfrac{1}{4}\Bbb{P}((X, Y) \in [-\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}}]^2) + \tfrac{1}{8}\Bbb{P}((X, Y) \in [-1,1]^2) \\
&= \Bbb{P}(X \in [0, \tfrac{1}{\sqrt{2}}])^2 + \frac{1}{2}\Bbb{P}(X \in [0, 1])^2 \\
&= \left( \Phi\left(\tfrac{1}{\sqrt{2}}\right) - \tfrac{1}{2}\right)^2 + \tfrac{1}{2}\left( \Phi(1)- \tfrac{1}{2}\right)^2.
\end{align*}
