50. Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive real numbers satisfying $x_{1}+x_{2}+\cdots+x_{n}=1$. Try to find the integer part of $E=x_{1}+$ $\frac{x_{2}}{\sqrt{1-x_{1}^{2}}}+\frac{x_{3}}{\sqrt{1-\left(x_{1}+x_{2}\right)^{2}}}+\cdots+\frac{x_{n}}{\sqrt{1-\left(x_{1}+x_{2}+\cdots+x_{n-1}\right)^{2}}}$.
50. Since $\sqrt{1-\left(x_{1}+x_{2}+\cdots+x_{i}\right)^{2}} \leqslant 1(1 \leqslant i \leqslant n)$, we have $E \geqslant x_{1}+ x_{2}+\cdots+x_{n}=1$

Let $\alpha_{1}=\frac{\pi}{2}, \alpha_{i}=\arccos \left(x_{1}+x_{2}+\cdots+x_{i-1}\right) \quad(2 \leqslant i \leqslant n)$, which means $x_{i}=\cos \alpha_{i+1}-\cos \alpha_{i}$, so
$$E=\frac{\cos \alpha_{2}-\cos \alpha_{1}}{\sin \alpha_{1}}+\frac{\cos \alpha_{3}-\cos \alpha_{2}}{\sin \alpha_{2}}+\cdots+\frac{\cos \alpha_{n+1}-\cos \alpha_{n}}{\sin \alpha_{n}}$$

Since $\alpha_{i+1}<\alpha_{i}, \sin x<x\left(0<x<\frac{\pi}{2}\right)$, we have $\frac{\cos \alpha_{i+1}-\cos \alpha_{i}}{\sin \alpha_{i}}=$ $\frac{2 \sin \frac{\alpha_{i+1}+\alpha_{i}}{2} \sin \frac{\alpha_{i}-\alpha_{i+1}}{2}}{\sin \alpha_{i}}<2 \sin \frac{\alpha_{i}-\alpha_{i+1}}{2}<2 \cdot \frac{\alpha_{i}-\alpha_{i+1}}{2}=\alpha_{i}-\alpha_{i+1}(1 \leqslant$ $i \leqslant n)$, thus $E<\left(\alpha_{1}-\alpha_{2}\right)+\left(\alpha_{2}-\alpha_{3}\right)+\cdots+\left(\alpha_{n}-\alpha_{n+1}\right)=\alpha_{1}-\alpha_{n+1}<$ $\alpha_{1}=\frac{\pi}{2}<2$, which means $1 \leqslant E<2,[E]=1$.