If 2 2 9 x 2 cos 2 x ( 1 + e x ) d x = then2 be equals [2025]

Question

If $\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{96 x^{2} \cos^{2} x}{(1 + e^{x})} d x = π (α π^{2} + β)$ , $α, β \in Z$ , then ${(α + β)}^{2}$ be equals [2025]

Smart Achievers · Accepted Answer

(3)

Let $I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{96 x^{2} \cos^{2} x}{(1 + e^{x})} d x$ ... (i)

$\Rightarrow I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{96 x^{2} \cos^{2} x}{(1 + e^{- x})} d x$ ... (ii)

$[∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x]$

Adding equations (i) and (ii), we get

$2 I = \int_{- π / 2}^{π / 2} \frac{96 x^{2} \cos^{2} x}{1 + e^{x}} d x + \int_{- π / 2}^{π / 2} \frac{e^{x} (96 x^{2} \cos^{2} x)}{e^{x} + 1} d x$

$= 2 \int_{0}^{π / 2} 96 x^{2} \cos^{2} x d x$

$\Rightarrow I = \int_{0}^{π / 2} 96 x^{2} \cos^{2} x d x$

$= 48 \int_{0}^{π / 2} x^{2} (1 + \cos 2 x) d x$

$= 48 [{[\frac{x^{3}}{3}]}_{0}^{π / 2} + \int_{0}^{π / 2} x^{2} \cos 2 x d x]$

$\Rightarrow I = 48 \cdot \frac{π^{3}}{24} + {[\frac{x^{2} \sin 2 x}{2}]}_{0}^{π / 2} - \int_{0}^{π / 2} x \sin 2 x$

$= 48 [\frac{π^{3}}{24} + {[\frac{x \cos 2 x}{2}]}_{0}^{π / 2} - \frac{1}{4} {[\sin 2 x]}_{0}^{π / 2}]$

$\Rightarrow I = 48 [\frac{π^{3}}{24} - \frac{π}{4}] = π (2 π^{2} - 12)$

So, $α = 2, β = - 12$

$∴ {(α + β)}^{2} = {(2 - 12)}^{2} = {(- 10)}^{2} = 100$ .

Solution

Q.
If $\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{96 x^{2} \cos^{2} x}{(1 + e^{x})} d x = π (α π^{2} + β)$ , $α, β \in Z$ , then ${(α + β)}^{2}$ be equals [2025]

1 64

2 144

3 100

4 196

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