Question

Evaluate: ${\int }_0^1\frac{xdx}{\sqrt{1+x^2}}$

Answer 1

${\int }_0^1\frac{xdx}{\sqrt{1+x^2}}$

$=\frac{1}{2}{\int }_0^1\frac{2xdx}{\sqrt{1+x^2}}$

Substituting $1+x^2=t\Rightarrow 2xdx=dt$,

The limits of the integration becomes
$x=0\Rightarrow t=1+0=1$
$x=1\Rightarrow t=1+1^2=2$

$\therefore \frac{1}{2}{\int }_0^1\frac{2xdx}{\sqrt{1+x^2}}=\frac{1}{2}{\int }_1^2\frac{dt}{\sqrt{t}}=\frac{1}{2}{\int }_1^2{t}^{-\frac{1}{2}}dt$

$=\frac{1}{2}{\left[\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]}_1^2$

$=\frac{1}{2}{\left[\frac{\sqrt{t}}{\frac{1}{2}}\right]}_1^2={\left[\sqrt{t}\right]}_1^2$

$=\sqrt{2}-\sqrt{1}$
$=\sqrt{2}-1$

Hence, the value of required integration is $\sqrt{2}-1$