Animated Solution for Mathematics - Definite Integration: $\frac{120}{{\pi }^3}{\int }_0^{\pi }\frac{x^2\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$ is equal to _______.

Welcome, future engineer. Today, we are not just solving an integral; we are embarking on a journey of mathematical elegance.

When you first look at the integral

it is natural to feel a surge of intimidation. It looks like a chaotic mess of powers, trigonometric functions, and a variable $x^2$ that seems to be in the way of everything. But in the world of JEE Advanced, intimidation is just a sign that you are about to learn something profound.

The first step in any complex integration problem is to observe the boundaries. We are integrating from $0$ to $\pi$.

In the realm of trigonometry, $\pi$ is a special number. It is the period of many functions, and it suggests symmetry. We decide to split our integral at the midpoint, $\pi /2$.

Because the denominator, ${\sin }^{4}x+{\cos }^{4}x$, is perfectly symmetric around $\pi /2$, we split the integral into:

We are not just splitting the area; we are creating two distinct zones to manipulate.

Now, we focus on the second integral, the one running from $\pi /2$ to $\pi$. We want it to look like the first one.

We use the substitution $x=\pi -t$. When $x=\pi /2$, $t=\pi /2$. When $x=\pi$, $t=0$. The differential $dx$ becomes $-dt$.

As we substitute, we observe the trigonometric terms: $\sin (\pi -t)$ remains $\sin t$, and $\cos (\pi -t)$ becomes $-\cos t$. The negative sign from the differential and the negative sign from the cosine term cancel out, and the limits flip back to $0$ to $\pi /2$.

Suddenly, the second integral is transformed into:

The limits now match the first integral perfectly.

This is the moment where the problem begins to collapse in our favor. We combine the two integrals into one:

Look at the numerator: $x^2-(\pi -x{)}^2$. If we expand $(\pi -x{)}^2$, we get ${\pi }^2-2\pi x+x^2$. Subtracting this from $x^2$ gives us $x^2-({\pi }^2-2\pi x+x^2)=2\pi x-{\pi }^2$.

The $x^2$ terms vanish! We are left with a much cleaner expression:

We can now split this into two integrals:

and

We solve $K_1$ first. By dividing the numerator and denominator by ${\cos }^{4}x$, we get:

Substituting $u={\tan }^{2}x$ turns this into a standard arctangent integral, yielding $\pi /4$.

Now for $J_1$. This is where King's Rule shines. By replacing $x$ with $(\pi /2-x)$, the $x$ term in the numerator is eliminated, and we find that $2J_1=\frac{\pi }{2}{K}_1$.

Thus, $J_1=\frac{\pi }{4}{K}_1=\frac{{\pi }^2}{16}$.

We plug these values back into our expression for $I$:

The question asks for $\frac{120}{{\pi }^3}|I|$. Substituting our result, we get:

You have navigated the complexity, respected the symmetry, and arrived at the truth. This is the essence of JEE Advanced mathematics: finding the simple path through the complex forest.