Question

The value of the integral \(\int _{-π/2}^{π/2}(\frac{x^2cosx}{1+e^x})dx\) is equal to \((\frac{π^2}{A})-B\). Then \((\frac{A}{B}) =\)

• A. \(-2\)
• B. \(2\)
• C. \(6\)
• D. \(-6\)

Hint:

Use the even function trick with 1/(1+e^x), then integrate by parts.

Answer 1

The Correct Option is B

Step 1: King property:
Write $I=\int_{-a}^a f(x)dx$ and also $I=\int_{-a}^a f(-x)dx$. Adding, $2I=\int_{-a}^a x^2\cos x\left[\frac1{1+e^x}+\frac{1}{1+e^{-x}}\right]dx$.

Step 2: Bracket:
The bracket equals $\dfrac{1}{1+e^x}+\dfrac{e^x}{e^x+1}=1$. So $2I=\int_{-\pi/2}^{\pi/2}x^2\cos x\,dx=2\int_0^{\pi/2}x^2\cos x\,dx$.

Step 3: Result:
$I=\dfrac{\pi^2}{4}-2$. Hence $A=4$, $B=2$, $A/B=2$.

Final Answer:
Both methods give I = pi^2/4 - 2. \[ \boxed{B} \]