Animated Solution for Mathematics - Definite Integration: The integral ${\int }_{1/4}^{3/4}\cos \left(2{\cot }^{-1}\sqrt{\frac{1-x}{1+x}}\right)dx$ is equal to

Welcome, warriors of JEE. Today, we face a problem that, at first glance, looks like a mathematical nightmare. We are tasked with evaluating the integral:

When you see a structure this complex—nested inverse trigonometric functions, square roots, and rational expressions—your brain might instinctively want to panic. But stop. Breathe. In the world of JEE Advanced, complexity is often just a mask for elegance. Our job is not to fight the monster, but to unmask it.

The first step in any battle is to simplify the terrain. We have a nested function: $\cos (2\theta )$, where $\theta ={\cot }^{-1}\sqrt{\frac{1-x}{1+x}}$.

By defining this substitution, we immediately strip away the intimidating outer layer. If $\theta ={\cot }^{-1}\sqrt{\frac{1-x}{1+x}}$, then by definition:

To make this even easier to handle, we square both sides: ${\cot }^2\theta =\frac{1-x}{1+x}$. Suddenly, the square root is gone. We have transformed a transcendental nightmare into a simple algebraic relationship. This is the first victory.

Now, we look at the integrand: $\cos (2\theta )$. We know $\cot \theta$, but we need $\cos (2\theta )$. Is there a bridge? Absolutely.

Recall the double-angle identity for cosine:

If we divide both the numerator and the denominator by ${\sin }^2\theta$, we get the beautiful identity:

This is the magic key. Let us substitute our value of ${\cot }^2\theta =\frac{1-x}{1+x}$ into this identity. The expression becomes:

Do not be intimidated by the complex fraction. Let us simplify the numerator:

Now, the denominator:

When we divide these two, the $(1+x)$ terms cancel out perfectly, leaving us with $\frac{-2x}{2}$, which is simply $-x$. The monster has vanished. We are left with the integral of $-x$.

What started as a terrifying integral has collapsed into:

This is a simple linear function. Geometrically, we are calculating the signed area under the line $y=-x$ between $x=1/4$ and $x=3/4$.

Using the power rule, the integral of $-x$ is $-\frac{x^2}{2}$. We evaluate this from $1/4$ to $3/4$:

Calculating the squares, we get $\frac{9}{16}-\frac{1}{16}=\frac{8}{16}=\frac{1}{2}$. Finally, we multiply by the constant outside:

And there it is. The final answer is $-1/4$. This problem teaches us a vital lesson: never judge a problem by its appearance. With the right substitution and a firm grasp of identities, even the most complex-looking expressions can collapse into something trivial.