Question

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

• A. $\frac{1}{\sqrt{2}} \, \tan^{-1} \sqrt{\frac{2}{3}}$
• B. $\frac{2}{\sqrt{2}} \tan^{-1} \left(\frac{3}{\sqrt{2}}\right)$
• C. $\frac{\sqrt{2}}{2} \tan^{-1} \left(\frac{3}{2}\right)$
• D. $\frac{\sqrt{2}}{2} \tan^{-1} \left(\frac{\sqrt{3}}{2}\right)$

Answer 1

Answer: (A)

Put $x = \sin\, \theta$
$\Rightarrow \,cos\,\theta\,d \theta=dx$
When $x = 0, \theta = 0$ and when $x = \frac{1}{2} , \theta = \frac{\pi}{6}$
$\therefore \; I = \int\limits^{\frac{\pi}{6}}_0 \frac{\cos \theta \, d \theta}{( 1 + \sin^2 \theta) \cos \theta}$
$ = \int\limits^{\frac{\pi}{6}}_0 \frac{\sec^2 \theta}{\sec^2 \theta + \tan^2 \theta} d \theta $
$ = \int\limits^{\frac{\pi}{6}}_0 \frac{\sec^2 \theta}{ 1 + (\sqrt{2} \tan \theta)^2} d \theta $
$ = \frac{1}{\sqrt{2}} \int\limits^{\frac{\pi}{6}}_0 \frac{\sqrt{2} \sec^2 \theta}{ 1 + (\sqrt{2} \tan \theta )^2} d \theta$
$ = \frac{1}{\sqrt{2}} \tan^{-1} (\sqrt{2} \tan \theta ) \Bigg|^{\frac{\pi}{6}}_0 = \frac{1}{\sqrt{2}} \tan^{-1} \left( \sqrt{\frac{2}{3}} \right)$