Question

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

Answer 1

Let $I={\int }_0^1\frac{dx}{x+\sqrt{1-x^2}}$
Put $x=\sin \theta$
$dx=\cos \theta d\theta$
$x=0\Rightarrow \theta =0$
$x=1\Rightarrow \theta =\pi /2$
Therefore, $I={\int }_0^{\pi /2}\frac{\cos \theta d\theta }{\sin \theta +\sqrt{1-{\sin }^2\theta }}$
$I={\int }_0^{\pi /2}\frac{\cos \theta d\theta }{\sin \theta +\cos \theta }$ ....(i)
$I={\int }_0^{\pi /2}\frac{\cos (\pi /2-\theta )d\theta }{\sin (\pi /2-\theta )+\cos (\pi /2-\theta )}$ (Using property ${\int }_a^{0}f\left(x\right)dx={\int }_0^{a}f(a-x)dx)$
$I={\int }_0^{\pi /2}\frac{\sin \theta d\theta }{\cos \theta +\sin \theta }$ .... (ii)
Adding (i) and (ii), we get
$2I={\int }_0^{\pi /2}\frac{(\sin \theta +\cos \theta )}{\sin \theta +\cos \theta )}d\theta$
$={\int }_0^{\pi /2}d\theta$
$=[\theta {]}_0^{\pi /2}$
$=\frac{\pi }{2}$
$\therefore I=\frac{\pi }{4}$