Question

\[\int_0^\frac{1}{2} \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx\]

Answer 1

\[\text{Let I } = \int_0^\frac{1}{2} \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx\]

Put

\[\int_0^\frac{1}{2} \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx\]

\[\therefore dx = \cos\theta d\theta\]

When `xrarr0, thetararr0`

 

When \[x \to \frac{1}{2}, \theta \to \frac{\pi}{6}\]

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

\[ = \int_0^\frac{\pi}{6} \theta\sin\theta d\theta\]

Applying integration by parts, we have

\[\left.I = \theta\left( - \cos\theta \right)\right|_0^\frac{\pi}{6} - \int_0^\frac{\pi}{6} 1 \times \left( - \cos\theta \right)d\theta\]
\[ = - \left( \frac{\pi}{6}\cos\frac{\pi}{6} - 0 \right) + \int_0^\frac{\pi}{6} \cos\theta d\theta\]
\[ = - \frac{\pi}{6} \times \frac{\sqrt{3}}{2} + \sin\theta_0^\frac{\pi}{6} \]
\[ = - \frac{\pi}{4\sqrt{3}} + \left( \sin\frac{\pi}{6} - \sin0 \right)\]
\[ = - \frac{\pi}{4\sqrt{3}} + \left( \frac{1}{2} - 0 \right)\]
\[ = \frac{1}{2} - \frac{\pi}{4\sqrt{3}}\]