Question

The value of the integral ${\int }_0^{\frac{1}{\sqrt{3}}}\frac{dx}{(1+x^2)\sqrt{1-x^2}}$

• A. $\frac{\pi }{2\sqrt{2}}$
• B. $\frac{\pi }{4\sqrt{2}}$
• C. $\frac{\pi }{8\sqrt{2}}$
• D. none of these

Answer 1

Answer: (B)

$\frac{\pi }{4\sqrt{2}}$

Orl putting $x=\sin \theta$ ,we get $dx=\cos \theta d\theta$
Integral (without limits)$=\int \frac{\cos \theta d\theta }{(1+{\sin }^2\theta )\left(\cos \theta \right)}$
$=\int \frac{d\theta }{1+{\sin }^2\theta }=\int \frac{cosec{}^2\theta d\theta }{2+{\cot }^2}$
$=\int \frac{-dt}{2+t^2}$ where $t=\cot \theta$
$=-\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{t}{\sqrt{2}}$
$=-\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{\cot \theta }{\sqrt{2}}$
$=-\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{1}{\sqrt{2}}\left(\frac{\sqrt{1-x^2}}{x}\right)$
$\therefore$ Definite integral $=-\frac{1}{\sqrt{2}}{\tan }^{-1}1+\frac{1}{\sqrt{2}}{\tan }^{-1}\infty$
$=-\frac{\pi }{4\sqrt{2}}+\frac{\pi }{2\sqrt{2}}=\frac{\pi }{4\sqrt{2}}$