JEE Mains · Maths · STD 12 - 7.2 definite integral
If \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos ^2 x}{\left(1+e^x\right)} \mathrm{d} x=\pi\left(\alpha \pi^2+\beta\right), \alpha, \beta \in \mathbb{Z}\), then \((\alpha+\beta)^2\) equals
- A 64
- B 196
- C 144
- D 100
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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