Question:

Here, $I = \int \sqrt{\frac{3-x}{3+x}} \cdot \sin^{-1}\left(\frac{1}{\sqrt{6}} \sqrt{3-x}\right) dx$<br>Put $x = 3\cos 2\theta \Rightarrow dx = -6\sin 2\theta d\theta$<br>$= \int \sqrt{\frac{3-3\cos 2\theta}{3+3\cos 2\theta}} \cdot \sin^{-1}\left(\frac{1}{\sqrt{6}} \sqrt{3-3\cos 2\theta}\right) (-6 \sin 2\theta) d heta$<br>$\therefore I = \int \frac{\sin \theta}{\cos \theta} \cdot \sin^{-1}(\sin \theta) \cdot (-6 \sin 2\theta) d\theta = -6 \int \theta \cdot (2 \sin^2 \theta) d\theta$<br>$= -6 \int \theta(1 - \cos 2\theta) d\theta = -6 \left\{ \frac{\theta^2}{2} - \int \theta \cos 2\theta d\theta \right\} + C$<br>$= -6 \left\{ \frac{\theta^2}{2} - \left( \theta \frac{\sin 2\theta}{2} - \int 1 \cdot \left( \frac{\sin 2\theta}{2} \right) d\theta \right) \right\} + C$<br>$= -3\theta^2 + 6 \left\{ \theta \frac{\sin 2\theta}{2} + \frac{\cos 2\theta}{4} \right\} + C$<br>$= \frac{1}{4} \left\{ -3 \left( \cos^{-1} \left( \frac{x}{3} \right) \right)^2 + 2\sqrt{9-x^2} \cdot \cos^{-1} \left( \frac{x}{3} \right) + 2x \right\} + C$