The integral $\underset{0}{\overset{1}{\int }}\frac{\mathrm{dx}}{\sqrt{1-\mathrm{x}}}$ is equal to _______

Given $\mathrm{I}=\underset{0}{\overset{1}{\int }}\frac{\mathrm{dx}}{\sqrt{1-\mathrm{x}}}$

Substituting x with sin²θ, we can write:

dx = 2sinθ cosθ dθ

The given integral can now be written as:

$\mathrm{I}=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{\sqrt{1-{\sin }^2\theta }}.2\sin \theta \cos \theta .\mathrm{d}\theta$

$I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{\cos \theta }.2\sin \theta \cos \theta .\mathrm{d}\theta$

$=2.{\left[-\cos \theta \right]}_0^{\frac{\pi }{2}}=\left(-2\right)\left(0-1\right)$

I = 2