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: Understanding the Concept:
For an even function \(h(x)\), \(\displaystyle\int_{-a}^{a}\frac{h(x)}{1+e^x}dx=\int_0^a h(x)dx\). Here \(h(x)=x^2\cos x\) is even.

Step 2: Reduce:
\[ I=\int_0^{\pi/2}x^2\cos x\,dx \]

Step 3: Integrate by parts:
\[ \int x^2\cos x\,dx=x^2\sin x-\int2x\sin x\,dx \]
\[ \int x\sin x\,dx=-x\cos x+\sin x \]

Step 4: Evaluate:
\[ I=\left[x^2\sin x+2x\cos x-2\sin x\right]_0^{\pi/2}=\frac{\pi^2}4+0-2=\frac{\pi^2}{4}-2 \]

Step 5: Read A and B:
\(A=4\) and \(B=2\), so \(\dfrac AB=2\). Option (B).

Final Answer:
The integral is pi^2/4 - 2, so A/B = 2. \[ \boxed{2} \]