Question

${\int }_0^{1/2}\frac{dx}{\sqrt{(1+x^2)\sqrt{1-x^2}}}$

Answer 1

Let I =${\int }_0^{1/2}\frac{dx}{\sqrt{(1+x^2)\sqrt{1-x^2}}}Putx=sin\theta \Rightarrow dx=cos\theta d\theta Asx\to 0,then\theta \to 0andx\to \frac{1}{2},then\theta \to \frac{\pi }{6}\therefore I={\int }_0^{\pi /6}\frac{cos\theta }{(1+si{n}^2\theta )cos\theta }d\theta ={\int }_0^{\pi /6}\frac{1}{1+si{n}^2\theta }d\theta ={\int }_0^{\pi /6}\frac{1}{co{s}^2\theta (se{c}^2\theta +ta{n}^2\theta )}={\int }_0^{\pi /6}\frac{se{c}^2\theta }{}se{c}^2\theta +ta{n}^2\theta d\theta ={\int }_0^{\pi /6}\frac{se{c}^2\theta }{1+2ta{n}^2\theta }d\theta Again,puttan\theta =t\Rightarrow se{c}^2\theta d\theta =dtAs\theta \to 0,thent\to 0and\theta \to \frac{\pi }{6},thent\to \frac{1}{\sqrt{3}}\therefore I={\int }_0^{1/\sqrt{3}}\frac{dt}{1+2t^2}=\frac{1}{2}{\int }_0^{1/\sqrt{3}}\frac{dt}{{\left(\frac{1}{\sqrt{2}}\right)}^2}+t^2=\frac{1}{2}.\frac{1}{1\sqrt{2}}{\left[ta{n}^{-1}\frac{\frac{t}{1}}{\sqrt{2}}\right]}_0^{1/\sqrt{3}}=\frac{1}{\sqrt{2}}\left[ta{n}^{-}1\sqrt{\frac{2}{3}-0}\right]=\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\sqrt{\frac{2}{3}}\right)$