Question

\[\int_0^\frac{1}{2} \frac{1}{\left( 1 + x^2 \right)\sqrt{1 - x^2}}dx\]

Answer 1

\[\text{Let I }= \int_0^\frac{1}{2} \frac{1}{\left( 1 + x^2 \right)\sqrt{1 - x^2}}dx\]

Put

\[x = \sin\theta\]

`therefore dx=costheta d theta`

When \[x \to 0, \theta \to 0\]

When `xrarr1/2, thetararrpi/6`

\[\therefore I = \int_0^\frac{\pi}{6} \frac{1}{\left( 1 + \sin^2 \theta \right)\cos\theta} \times \cos\theta d\theta\]
\[ = \int_0^\frac{\pi}{6} \frac{1}{1 + \sin^2 \theta}d\theta\]

Dividing numerator and denominator by `cos^2theta, `we have

\[I = \int_0^\frac{\pi}{6} \frac{\sec^2 \theta}{\sec^2 \theta + \tan^2 \theta}d\theta\]
\[ = \int_0^\frac{\pi}{6} \frac{\sec^2 \theta}{1 + 2 \tan^2 \theta}d\theta\]

Now, put `tantheta = u`

`therefore sec^2thetad theta=du`

When `thetararr0, u rarr0`

When \[\theta \to \frac{\pi}{6}, u \to \frac{1}{\sqrt{3}}\]

\[\therefore I = \int_0^\frac{1}{\sqrt{3}} \frac{du}{1 + 2 u^2}\]
\[ = \int_0^\frac{1}{\sqrt{3}} \frac{du}{1 + \left( \sqrt{2}u \right)^2}\]
\[ = \left.\frac{\tan^{- 1} \sqrt{2}u}{\sqrt{2}}\right|_0^\frac{1}{\sqrt{3}} \]
\[ = \frac{1}{\sqrt{2}}\left( \tan^{- 1} \frac{\sqrt{2}}{\sqrt{3}} - 0 \right)\]
\[ = \frac{1}{\sqrt{2}} \tan^{- 1} \sqrt{\frac{2}{3}}\]