Evaluate $\int \sqrt{\frac{a + x}{a - x}} \, dx$

$-a \left[ \cos^{-1} \left( \frac{x}{a} \right) + \frac{\sqrt{a^2 - x^2}}{a} \right] + C$

The correct answer is Option (3) → $-a \left[ \cos^{-1} \left( \frac{x}{a} \right) + \frac{\sqrt{a^2 - x^2}}{a} \right] + C$

Let $I = \int \sqrt{\frac{a + x}{a - x}} \, dx$

Put $x = a \cos 2\theta ⇒dx = -a \cdot \sin 2\theta \cdot 2 \cdot d\theta$

$∴I = -2 \int \sqrt{\frac{a + a \cos 2\theta}{a - a \cos 2\theta}} \cdot a \sin 2\theta d\theta$

$= -2a \int \sqrt{\frac{1 + \cos 2\theta}{1 - \cos 2\theta}} \sin 2\theta d\theta = -2a \int \sqrt{\frac{2\cos^2 \theta}{2\sin^2 \theta}} \sin 2\theta d\theta$

$= -2a \int \cot \theta \cdot \sin 2\theta d\theta = -2a \int \frac{\cos \theta}{\sin \theta} \cdot 2 \sin \theta \cdot \cos \theta \, d\theta$    $[∵\sin 2\theta = 2 \sin \theta \cos \theta]$

$= -4a \int \cos^2 \theta d\theta = -2a \int (1 + \cos 2\theta) \, d\theta \quad [∵ \cos 2\theta = 2 \cos^2 \theta - 1]$

$= -2a \left[ \theta + \frac{\sin 2\theta}{2} \right] + C$

$∵\cos 2\theta = \frac{x}{a} ⇒2\theta = \cos^{-1} \frac{x}{a} ⇒\theta = \frac{1}{2} \cos^{-1} \frac{x}{a}$

Now, $\sin 2\theta = \frac{\sqrt{a^2 - x^2}}{a}$

$∴I = -2a \left[ \frac{1}{2} \cos^{-1} \frac{x}{a} + \frac{1}{2} \frac{\sqrt{a^2 - x^2}}{a} \right] + C$

$= -a \left[ \cos^{-1} \left( \frac{x}{a} \right) + \frac{\sqrt{a^2 - x^2}}{a} \right] + C$