The integral $$\int_{1/4}^{3/4} \cos\left(2\cot^{-1}\sqrt{\frac{1-x}{1+x}}\right) dx$$ is equal to

We need to evaluate $$\int_{1/4}^{3/4} \cos\left(2\cot^{-1}\sqrt{\frac{1-x}{1+x}}\right)dx$$.

First, we simplify the inner expression by setting $$x = \cos\theta$$, where $$\theta \in [0, \pi]$$. Then

$$\frac{1-x}{1+x} = \frac{1-\cos\theta}{1+\cos\theta}$$

Using the half-angle identities $$1-\cos\theta = 2\sin^2(\theta/2)$$ and $$1+\cos\theta = 2\cos^2(\theta/2)$$, we obtain

$$\frac{1-\cos\theta}{1+\cos\theta} = \frac{2\sin^2(\theta/2)}{2\cos^2(\theta/2)} = \tan^2(\theta/2)$$

Hence, $$\sqrt{\frac{1-x}{1+x}} = \tan(\theta/2) \quad \text{(positive for } \theta \in (0,\pi)\text{)}$$, and thus

$$\cot^{-1}(\tan(\theta/2)) = \frac{\pi}{2} - \tan^{-1}(\tan(\theta/2)) = \frac{\pi}{2} - \frac{\theta}{2}$$

Since $$\theta/2 \in (0, \pi/2)$$ for the given range, it follows that $$\tan^{-1}(\tan(\theta/2)) = \theta/2$$.

It follows that

$$\cos\left(2\cot^{-1}\sqrt{\frac{1-x}{1+x}}\right) = \cos\left(2 \cdot \left(\frac{\pi}{2} - \frac{\theta}{2}\right)\right) = \cos(\pi - \theta) = -\cos\theta = -x$$

Substituting into the integral gives

$$\int_{1/4}^{3/4} (-x) \, dx = -\frac{x^2}{2}\Bigg|_{1/4}^{3/4} = -\frac{1}{2}\left(\frac{9}{16} - \frac{1}{16}\right) = -\frac{1}{2} \times \frac{8}{16} = -\frac{1}{2} \times \frac{1}{2} = -\frac{1}{4}$$

The correct answer is Option 3: $$-\frac{1}{4}$$.