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JEE Mains · Maths · STD 12 - 7.2 definite integral

જો \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos ^2 x}{\left(1+e^x\right)} \mathrm{d} x=\pi\left(\alpha \pi^2+\beta\right), \alpha, \beta \in \mathbb{Z}\) હોય, તો \((\alpha+\beta)^2\) નું મૂલ્ય કેટલું થશે?

  1. A 64
  2. B 196
  3. C 144
  4. D 100
Verified Solution

Answer & Solution

Correct Answer

(D) 100

Step-by-step Solution

Detailed explanation

\begin{aligned} & I=\int_0^{\frac{\pi}{2}} \frac{96 x^2 \cos ^2 x}{1+e^x} d x \\ & 2 I=2 \int_0^{\frac{\pi}{2}} 96 x^2 \cos ^2 x d x \\ & I=96 \int_0^{\frac{\pi}{2}} x^2 \cos ^2 x d x \\ & =48 \int_0^{\frac{\pi}{2}} x^2(1+\cos 2 x) d x \\ & =2 \pi^2+48(0-0)-48…

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