How do you test the improper integral int x/sqrt(1-x^2)dx from [0,1] and evaluate if possible?

Mar 31, 2017

$1$

Explanation:

Making $x = \sin y$ with $\mathrm{dx} = \cos y \mathrm{dy}$ and substituting into the integral

$I = \int \frac{x \mathrm{dx}}{\sqrt{1 - {x}^{2}}} \equiv \int \left(\sin \frac{y}{\cos} y\right) \cos y \mathrm{dy} = \int \sin y \mathrm{dy} = - \cos y$

The integration limits are changed to

$x = 0 \to y = 0$
$x = 1 \to y \to \frac{\pi}{2}$

so

$I = - \cos \left(\frac{\pi}{2}\right) - \left(- \cos \left(0\right)\right) = 1$

Mar 31, 2017

${\int}_{0}^{1} \frac{x}{\sqrt{1 - {x}^{2}}} \mathrm{dx} = 1$

Explanation:

We can also use the substitution $u = 1 - {x}^{2}$ implying that $\mathrm{du} = - 2 x \mathrm{dx}$.

$\int \frac{x}{\sqrt{1 - {x}^{2}}} \mathrm{dx} = - \frac{1}{2} \int \frac{- 2 x \mathrm{dx}}{\sqrt{1 - {x}^{2}}} = - \frac{1}{2} \int {u}^{- \frac{1}{2}} \mathrm{du}$

$= - \frac{1}{2} \left({u}^{\frac{1}{2}} / \left(\frac{1}{2}\right)\right) = - \sqrt{u} = - \sqrt{1 - {x}^{2}} + C$

So the integral with bounds is:

${\int}_{0}^{1} \frac{x}{\sqrt{1 - {x}^{2}}} \mathrm{dx} = {\left[- \sqrt{1 - {x}^{2}}\right]}_{0}^{1} = - \sqrt{1 - 1} - \left(- \sqrt{1 - 0}\right) = 1$