Question:

Let $I = \int \sqrt{\frac{x-3}{5-x}}dx$<br>Put $x = 3\cos^2\theta + 5\sin^2\theta$<br>$\Rightarrow x - 3 = 2\sin^2\theta, 5 - x = 2\cos^2\theta$<br>and $dx = 4\sin\theta\cos\theta d\theta$<br>$\therefore I = \int \frac{\sin\theta}{\cos\theta} \times 4\sin\theta\cos\theta d\theta = 4\int \sin^2\theta d\theta$<br>$= 2\int (1 - \cos 2\theta) d\theta = 2\left[\theta - \frac{\sin 2\theta}{2}\right] + C$<br>$= 2\tan^{-1}\sqrt{\frac{x-3}{5-x}} - \sqrt{(x-3)(5-x)} + C$