The value of the integral (48/π⁴) ∫[0 to π] x² (3πx²/2 - x³) (sin x / (1 + cos² x)) dx is equal to _

Q: Wait, why do we use King's Property here? Can't we just integrate by parts?

Integration by parts would lead to a nightmare of nested products and powers. King's Property is specifically designed for definite integrals with symmetric limits or trigonometric functions that repeat under reflection. By replacing $x$ with $\pi - x$, we exploit the symmetry of the interval $[0, \pi]$, which causes the complex polynomial terms to collapse into a much simpler form.

Q: Why does the polynomial $S(x) = P(x) + P(\pi - x)$ simplify so much?

It is not a coincidence! The polynomial was constructed specifically to be symmetric about the midpoint $\frac{\pi}{2}$. When you add $P(x)$ and $P(\pi - x)$, the odd-powered terms that break symmetry cancel out, leaving only terms that are symmetric about $\frac{\pi}{2}$. This is a common pattern in JEE Advanced problems to reward students who look for structure before diving into brute-force calculation.

Q: Why shift the variable to $t = x - \frac{\pi}{2}$?

Shifting the variable is a powerful technique to center the interval of integration. By moving from $[0, \pi]$ to $[-\frac{\pi}{2}, \frac{\pi}{2}]$, we can use the properties of even and odd functions. In our case, the polynomial becomes an even function of $t$, and the trigonometric part transforms into a form that is much easier to handle, effectively isolating the constant factor we need.

Question

The value of the integral $\frac{48}{\pi^4} \int_{0}^{\pi} x^2 \left( \frac{3\pi x^2}{2} - x^3 \right) \frac{\sin x}{1 + \cos^2 x} dx$ is equal to ______.

Sigma Percentile · Accepted Answer

### The Beauty of Symmetry: Unlocking the Integral

Welcome, future engineer! Today, we are going to embark on a journey to solve an integral that, at first glance, looks like a chaotic mess of polynomials and trigonometry.

In the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back that mask together.

### Phase 1: The King's Property

Consider the integral:

$$I = \frac{48}{\pi^4} \int_{0}^{\pi} x^2 \left( \frac{3\pi x^2}{2} - x^3 ight) \frac{\sin x}{1 + \cos^2 x} dx$$

The limits are $0$ to $\pi$. Whenever you see these specific limits, your intuition should immediately scream: **King's Property!**

This property, $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$, is our most powerful tool for handling trigonometric functions on this interval. By replacing $x$ with $\pi - x$, we reflect the function across the midpoint $\frac{\pi}{2}$.

Since $\sin(\pi - x) = \sin x$ and $\cos^2(\pi - x) = \cos^2 x$, the trigonometric heart of the integral remains unchanged.

### Phase 2: The Polynomial Dance

Let us define the polynomial part as $P(x) = \frac{3\pi x^4}{2} - x^5$.

When we apply King's Property and add the original integral $I$ to the transformed integral $I$, we obtain:

$$2I = \frac{48}{\pi^4} \int_{0}^{\pi} [P(x) + P(\pi - x)] \frac{\sin x}{1 + \cos^2 x} dx$$

Now, the magic happens. We calculate $S(x) = P(x) + P(\pi - x)$. If you expand this carefully, you will see the higher-order terms vanish, leaving a polynomial symmetric about $\frac{\pi}{2}$.

### Phase 3: The Shift to Simplicity

To fully exploit this symmetry, let us shift our origin. We set $x = t + \frac{\pi}{2}$, which changes our limits to $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

The polynomial $S(t + \frac{\pi}{2})$ simplifies to:

$$S(t) = -2\pi t^4 + 2\pi^3 t^2 + \frac{\pi^5}{8}$$

This is an even function. When we integrate an even function multiplied by a symmetric trigonometric term over a symmetric interval, the odd components vanish. The entire polynomial part effectively reduces to a constant factor of $\frac{\pi^3}{2}$.

### Phase 4: The Final Victory

After the reduction, we are left with a standard integral:

$$I = \frac{12}{\pi} \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} dx$$

We use the substitution $u = \cos x$, which gives $du = -\sin x dx$. The limits change from $1$ to $-1$, and the negative sign flips them back to $-1$ to $1$.

We are left with:

$$I = \frac{12}{\pi} \int_{-1}^{1} \frac{du}{1 + u^2}$$

The integral of $\frac{1}{1+u^2}$ is simply $	an^{-1} u$. Evaluating this from $-1$ to $1$ gives us $\frac{\pi}{4} - (-\frac{\pi}{4}) = \frac{\pi}{2}$.

Finally, multiplying by $\frac{12}{\pi}$ yields our beautiful answer: **6**.

Animated Solution for Mathematics - Definite Integration: The value of the integral $\frac{48}{π ^{4}} \int_{0}^{π} x^{2} (\frac{3 π x ^{2}}{2} - x^{3}) \frac{s i n x}{1 + c o s ^{2} x} d x$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Defining the Integral $I$

Applying King's Property

Combining the Integrals

Simplifying the Polynomial Sum

Shifting the Variable

The Reduction Insight

Standard Integration Form

Final Calculation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Beauty of Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

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