Integral Calculus – Definite Integration King’s Rule – JEE Main 28 Jan 2025 Shift 1

Solution:

Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos^2 x}{1 + e^x} dx$.

Using the property $\int_{-a}^{a} f(x) dx = \int_{0}^{a} [f(x) + f(-x)] dx$:
$$ f(x) + f(-x) = \frac{96 x^2 \cos^2 x}{1 + e^x} + \frac{96 (-x)^2 \cos^2 (-x)}{1 + e^{-x}} $$
$$ = 96 x^2 \cos^2 x \left( \frac{1}{1+e^x} + \frac{1}{1+\frac{1}{e^x}} \right) $$
$$ = 96 x^2 \cos^2 x \left( \frac{1}{1+e^x} + \frac{e^x}{e^x+1} \right) = 96 x^2 \cos^2 x $$

The integral reduces to:
$$ I = \int_{0}^{\frac{\pi}{2}} 96 x^2 \cos^2 x \, dx = 96 \int_{0}^{\frac{\pi}{2}} x^2 \left(\frac{1 + \cos 2x}{2}\right) dx $$
$$ I = 48 \int_{0}^{\frac{\pi}{2}} x^2 dx + 48 \int_{0}^{\frac{\pi}{2}} x^2 \cos 2x \, dx $$

Solving the first part:
$$ 48 \left[ \frac{x^3}{3} \right]_{0}^{\frac{\pi}{2}} = 16 \left(\frac{\pi}{2}\right)^3 = 16 \cdot \frac{\pi^3}{8} = 2\pi^3 $$

Solving the second part using Integration by Parts ($\int u v’ = uv – \int u’v$):
$$ \int_{0}^{\frac{\pi}{2}} x^2 \cos 2x \, dx = \left[ x^2 \frac{\sin 2x}{2} \right]_{0}^{\frac{\pi}{2}} – \int_{0}^{\frac{\pi}{2}} 2x \frac{\sin 2x}{2} dx $$
$$ = \left[ 0 – 0 \right] – \int_{0}^{\frac{\pi}{2}} x \sin 2x \, dx $$
$$ = – \left( \left[ -x \frac{\cos 2x}{2} \right]_{0}^{\frac{\pi}{2}} – \int_{0}^{\frac{\pi}{2}} 1 \cdot \left(-\frac{\cos 2x}{2}\right) dx \right) $$
$$ = – \left( \left[ -\frac{\pi}{2} \cdot \frac{(-1)}{2} – 0 \right] + \frac{1}{2} \left[ \frac{\sin 2x}{2} \right]_{0}^{\frac{\pi}{2}} \right) $$
$$ = – \left( \frac{\pi}{4} + 0 \right) = -\frac{\pi}{4} $$

Combining the results:
$$ I = 2\pi^3 + 48 \left(-\frac{\pi}{4}\right) = 2\pi^3 – 12\pi $$
$$ I = \pi(2\pi^2 – 12) $$

Comparing with $\pi(\alpha\pi^2 + \beta)$:
$$ \alpha = 2, \quad \beta = -12 $$
$$ (\alpha + \beta)^2 = (2 – 12)^2 = (-10)^2 = 100 $$

Ans. (3)