$P(X_1 > 0 \mid X_1 + X_2 > 0)$ for IID $X_1, X_2 \sim \mathcal{N}(0,1)$ Given IID $X_1, X_2 \sim \mathcal{N}(0,1)$, we want to determine $P(X_1 > 0 \mid X_1 + X_2 > 0)$.
This is what I think is the approach for this problem:
\begin{align}
    P(X_1>0\mid X_1 + X_2 > 0) = P(X_1 > 0 \mid X_1 > -X_2) \\
    P(X_1 \leq x \mid X_1 > (-X_2=x_2)) = \int_{-x_2}^x \frac{1}{\sqrt{2\pi}}\exp(-\frac{1}{2}x_1^2) \, dx_1
\end{align}
I integrate the above to get $P(X_1 \leq x \mid  X_1 > (-X_2=x_2))$, and then I integrate $P(X_1 \leq x_1\mid X_1 > (-X_2=x_2))$ over all $x_2$:
$$
P(X_1 \leq x_1\mid X_1 > X_2) = \int_{-\infty}^\infty P(X_1 \leq x_1\mid X_1 > (-X_2=x_2)) \, dx_2
$$
Is this a correct approach? I feel that there is something simpler than this.
 A: The formula for conditional probability is $$P(A | B) = \frac{P(A \cap B)}{P(B)}$$
This then means that the answer is $$\frac{P(X_1 > 0 \cap X_1 + X_2 > 0)}{P(X_1 + X_2 > 0)}$$
Clearly, the denominator is $1/2$. Then for the numerator, the integral is $$\iint_{R} p(x_1) p(x_2) \,dx_1 \,dx_2$$
where $R$ is the region satisfying $x_1>0, x_1 + x_2 > 0$ and $p(x)$ is the equation for the probability distribution.
Plugging everything in, the integral comes out to $\frac{3}{8}$, which means the final probability comes out to $$\frac{\frac{3}{8}}{\frac{1}{2}} = \frac{3}{4}$$
A: 
Is this a correct approach? I feel that there is something simpler than this.

There is no integration required when you use a graphical approach
When 3D-plotting the joint probability density against the $X_1,X_2$ plane, you obtain a radially-symmetrical bell-shaped 'pie' stretching infinitely but whose total volume measure is $1$.   So you can readily evaluate the probability of slices of this 'pie'.
$$\begin{align}\mathsf P(X_1>0\mid X_1>-X_2)&=\dfrac{\mathsf P(X_1>\max\{0,-X_2\})}{\mathsf P(X_1>-X_2)}\\[1ex]&=\dfrac{\mathsf P(X_1>-X_2>0)+\mathsf P(X_1>0\geq -X_2)}{\mathsf P(X_1>-X_2)}\\&~~\vdots\end{align}$$
A: In polar coordinates, divide the $(x_1,x_2)$-plane into eight sectors: $0 \le \theta < \pi/4,$ then $\pi/4\le \theta < 2\pi/4, $ and so on, up to $7\pi/4\le\theta < 8\pi/4.$
All eight of these have equal probability of being where the pair $(X_1,X_2)$ is, by symmetry.
Call the random angle $\Theta,$ so that $(X_1,X_2) = (\cos\Theta,\sin\Theta)\cdot \sqrt{X_1^2 + X_2^2}.$
Then the event $X_1+X_2>0$ is the same as $-\pi/4<\Theta< 3\pi/4$ (here we treat $-\pi/4<\Theta<0$ as being the same thing as $7\pi/4<\Theta< 0\text{ or } 0\le\Theta<3\pi/4,$ i.e. in effect we're adding modulo $2\pi.$
And the event $X_1>0$ is the same as $-\pi/2<\Theta< \pi/2.$
In effect $\Pr( X_1 > 0 \mid X_1+X_2>0)$ is the probability that $\Theta$ is in sector $7,8,1, \text{ or }2$ given that $\Theta$ is in sector $8,1,2,\text{ or } 3.$ And all eight sectors are equally probable.
So it's $3/4.$
