If I = ∫96x²cos²x/(1+eˣ)dx = π(απ²+β), Find (α+β)²

JEE Main 2025 — 28 January, Morning Shift. Previous Year Question.

Problem

If $I = \displaystyle\int_{-\pi/2}^{\pi/2} \dfrac{96x^2\cos^2 x}{1+e^x}\,dx = \pi(\alpha\pi^2+\beta)$, then $(\alpha+\beta)^2$ is:

Key insight. The integrand isn’t even or odd on its own, so the obvious symmetry shortcuts don’t apply directly. But adding the integral of $f(x)$ to the integral of $f(-x)$ — the “King’s property,” $\int_{-a}^{a} f(x)\,dx = \int_{-a}^{a} f(a + (-a) – x)\,dx$ — makes the awkward $\dfrac{1}{1+e^x}$ denominator cancel out entirely, leaving a purely even function that’s much easier to integrate.

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Approach

Since the limits are symmetric about zero, it’s always worth checking whether King’s property — replacing $x$ with $-x$ inside the integral — simplifies things, even when the function isn’t obviously even or odd. Here, substituting $x \to -x$ turns the denominator $1+e^x$ into $1+e^{-x} = \dfrac{e^x+1}{e^x} = \dfrac{e^x(1+e^{-x})}{e^x}$, effectively flipping which term carries the $e^x$ factor. Adding the original integral to this transformed version cancels the exponential entirely, leaving behind a clean, even trigonometric-polynomial integral that integration by parts can finish.

Solution

Step 1 — Apply King’s property

Let $f(x) = \dfrac{x^2\cos^2 x}{1+e^x}$. Substituting $x \to -x$:

$$f(-x) = \frac{x^2\cos^2 x}{1+e^{-x}} = \frac{e^x \cdot x^2\cos^2 x}{e^x+1}$$

Step 2 — Add the original and the substituted integrand

$$f(x) + f(-x) = \frac{x^2\cos^2 x}{1+e^x} + \frac{e^x x^2\cos^2 x}{1+e^x} = \frac{x^2\cos^2 x(1+e^x)}{1+e^x} = x^2\cos^2 x$$

Since $\displaystyle\int_{-a}^{a} f(x)\,dx = \int_{-a}^{a} f(-x)\,dx$ (a direct consequence of King’s property), doubling the original integral gives:

$$2I = 96\int_{-\pi/2}^{\pi/2} x^2\cos^2 x\,dx \implies I = 48\int_{-\pi/2}^{\pi/2} x^2\cos^2 x\,dx$$

Since $x^2\cos^2 x$ is now a genuinely even function, this becomes:

$$I = 96\int_{0}^{\pi/2} x^2\cos^2 x\,dx$$

Step 3 — Rewrite $\cos^2 x$ and split the integral

Using $\cos^2 x = \dfrac{1+\cos 2x}{2}$:

$$I = 96\left[\frac{1}{2}\int_0^{\pi/2} x^2\,dx + \frac{1}{2}\int_0^{\pi/2} x^2\cos 2x\,dx\right] = 48\int_0^{\pi/2} x^2\,dx + 48\int_0^{\pi/2} x^2\cos 2x\,dx$$

The first piece is straightforward:

$$48\int_0^{\pi/2} x^2\,dx = 48 \cdot \frac{x^3}{3}\Big|_0^{\pi/2} = 16\cdot\frac{\pi^3}{8} = \frac{2\pi^3}{1} \cdot \frac{1}{1}= \frac{\pi^3}{1}\cdot 2$$

Carrying the arithmetic through carefully: $\dfrac{(\pi/2)^3}{3} = \dfrac{\pi^3}{24}$, so $48\times\dfrac{\pi^3}{24} = 2\pi^3$.

Step 4 — Integrate $x^2\cos 2x$ by parts (twice)

$$\int x^2\cos 2x\,dx = x^2\cdot\frac{\sin 2x}{2} – \int 2x\cdot\frac{\sin 2x}{2}\,dx = \frac{x^2\sin 2x}{2} – \int x\sin 2x\,dx$$

Integrating $\int x\sin 2x\,dx$ by parts again:

$$\int x\sin 2x\,dx = -\frac{x\cos 2x}{2} + \int \frac{\cos 2x}{2}\,dx = -\frac{x\cos 2x}{2} + \frac{\sin 2x}{4}$$

So:

$$\int x^2\cos 2x\,dx = \frac{x^2\sin 2x}{2} + \frac{x\cos 2x}{2} – \frac{\sin 2x}{4}$$

Evaluating from $0$ to $\pi/2$: at $x=\pi/2$, $\sin 2x = \sin\pi = 0$ and $\cos 2x = \cos\pi = -1$, so only the middle term survives: $\dfrac{(\pi/2)(-1)}{2} = -\dfrac{\pi}{4}$. At $x=0$, every term is $0$. So:

$$\int_0^{\pi/2} x^2\cos 2x\,dx = -\frac{\pi}{4}$$

Multiplying by $48$: $48\times\left(-\dfrac{\pi}{4}\right) = -12\pi$.

Step 5 — Combine and identify $\alpha$, $\beta$

$$I = 2\pi^3 – 12\pi = \pi(2\pi^2 – 12)$$

Comparing with $I = \pi(\alpha\pi^2+\beta)$: $\alpha = 2$, $\beta = -12$.

Step 6 — Compute $(\alpha+\beta)^2$

$$(\alpha+\beta)^2 = (2-12)^2 = (-10)^2 = 100$$

Answer

Common mistakes

• Missing that the function is neither even nor odd on its own, and trying to argue the integral is zero — the $\dfrac{1}{1+e^x}$ factor breaks that symmetry, which is exactly why King’s property (rather than a simple even/odd check) is needed here.
• Sign errors when evaluating the integration-by-parts result at the limits — $\cos\pi = -1$ is easy to mishandle, and a dropped negative sign here flips the final value of $\alpha$ or $\beta$ entirely.

Practise next

• Evaluate $\displaystyle\int_{-\pi/4}^{\pi/4} \dfrac{x^2}{1+e^x}\,dx$ using the same King’s property technique, to practise recognising when a denominator like $1+e^x$ signals this approach.