jee mains maths JEE (Main)-2025 (Online) Phase-1 28/01/2025 Morning

1. If I = ∫_−π/2^{π/2 96x2 cos2x}⁄_{1 + e^x} dx = π(απ² + β), then (α + β)² is

(1) 100
(2) 144
(3) 169
(4) 196

Solution to JEE Integral Question

The given integral is:

I = ∫_-π/2^π/2 (96x² cos² x)/(1 + e^x) dx = π(απ² + β)

Step 1: Use Symmetry Property

Let f(x) = (96x² cos² x)/(1 + e^x)

When we replace x with -x:

f(-x) = (96x² cos² x)/(1 + e^-x)

Since 1 + e^-x = (1 + e^x)/e^x, we get:

f(-x) = (96x² cos² x · e^x)/(1 + e^x)

Step 2: Using Properties of Definite Integrals

Over symmetric limits, we can write:

I = ∫_-π/2^π/2 (96x² cos² x)/(1 + e^x) dx = ∫_-π/2^π/2 (96x² cos² x · e^x)/(1 + e^x) dx

Adding these equal integrals:

2I = ∫_-π/2^π/2 96x² cos² x dx

Therefore: I = 48 ∫_-π/2^π/2 x² cos² x dx

Step 3: Using Trigonometric Identity

Using cos² x = (1 + cos 2x)/2:

I = 48 ∫_-π/2^π/2 x² · (1 + cos 2x)/2 dx

I = 24 ∫_-π/2^π/2 x² dx + 24 ∫_-π/2^π/2 x² cos 2x dx

Step 4: Evaluating the First Integral

∫_-π/2^π/2 x² dx = [x³/3]_-π/2^π/2 = (π/2)³/3 - (-π/2)³/3 = π³/12

Step 5: Evaluating the Second Integral Using Integration by Parts

Using integration by parts twice:

∫_-π/2^π/2 x² cos 2x dx = -π/2

Step 6: Finding the Value of the Original Integral

I = 24 · π³/12 + 24 · (-π/2) = 2π³ - 12π

Step 7: Finding α and β

Comparing with the given form I = π(απ² + β):

2π³ - 12π = απ³ + βπ

Therefore: α = 2 and β = -12

So α + β = 2 + (-12) = -10

And (α + β)² = (-10)² = 100

The answer is (1) 100.