Animated Solution for Mathematics - Definite Integration: Evaluate the definite integral : ${\int }_{-1/\sqrt{3}}^{1/\sqrt{3}}\left(\frac{x^4}{1-x^4}\right){\cos }^{-1}\left(\frac{2x}{1+x^2}\right)dx$

Cracking the Integral

Welcome, warriors of JEE Advanced. Today, we stand before an integral that, at first glance, looks like a chaotic mess. We have an integrand involving a rational function multiplied by an inverse trigonometric function:

It is intimidating, yes. But in the world of competitive mathematics, intimidation is often just a mask for elegance. Let us peel back that mask.

The Symmetry Insight

The first thing you must train your eyes to see is the domain. We are integrating from $-a$ to $a$, where $a=1/\sqrt{3}$. This is not a coincidence; it is a massive, flashing neon sign from the examiner.

Whenever you see symmetric limits, your brain should immediately reach for the property:

This property is the 'King's Property' of symmetric intervals. It allows us to fold the negative half of the domain onto the positive half, effectively combining the function's behavior at $x$ and $-x$.

The Inverse Trig Trap

Now, let us define our function $f(x)=\left(\frac{x^4}{1-x^4}\right){\cos }^{-1}\left(\frac{2x}{1+x^2}\right)$. To use our property, we need to evaluate $f(-x)$.

Substituting $-x$ into the rational part is easy: since the power is $4$ (an even power), $(-x{)}^4$ simply becomes $x^4$. The denominator remains $1-x^4$.

The real action happens in the inverse cosine term. We get ${\cos }^{-1}\left(\frac{2(-x)}{1+(-x{)}^2}\right)$, which is ${\cos }^{-1}\left(-\frac{2x}{1+x^2}\right)$.

Here is where the JEE examiners test your precision. We must use the identity ${\cos }^{-1}(-u)=\pi -{\cos }^{-1}(u)$.

Applying this, our $f(-x)$ transforms into:

Do you see the beauty? When we add $f(x)$ and $f(-x)$, the ${\cos }^{-1}$ terms cancel out perfectly! We are left with $\pi \frac{x^4}{1-x^4}$.

The nightmare has vanished, replaced by a clean, manageable integral:

The Algebraic Cleanup

We are not done yet, but the path is clear. We have $\frac{x^4}{1-x^4}$. Since the degree of the numerator equals the degree of the denominator, we perform a simple algebraic adjustment:

Now, we focus on $\frac{1}{1-x^4}$. Using the difference of squares, we factor the denominator into $(1-x^2)(1+x^2)$.

Through partial fractions, this splits into:

The Final Integration

Now, we integrate term by term. The integral of $-1$ is $-x$. The integral of $\frac{1}{1-x^2}$ is $\frac{1}{2}\ln \left|\frac{1+x}{1-x}\right|$, and the integral of $\frac{1}{1+x^2}$ is ${\tan }^{-1}(x)$.

Combining these with our constants, we evaluate from $0$ to $1/\sqrt{3}$. Substituting the upper limit $1/\sqrt{3}$ yields a logarithmic term $\ln \left(\frac{1+1/\sqrt{3}}{1-1/\sqrt{3}}\right)$, which simplifies beautifully to $\ln (2+\sqrt{3})$ after rationalization.

The ${\tan }^{-1}(1/\sqrt{3})$ gives us $\pi /6$. After multiplying by the $\pi$ we factored out earlier and simplifying the constants, we arrive at our final, elegant result:

You have conquered the beast. Remember, in JEE Advanced, it is rarely about brute force; it is about recognizing the symmetry and applying the right identity at the right moment. Keep practicing, and keep falling in love with the process.