1. Let $I=\int_{-\pi/2}^{\pi/2} \frac{x^2\cos x}{1+e^x}\,dx.$

We use the standard symmetry trick by replacing $x$ with $-x$.

2. Define $I=\int_{-\pi/2}^{\pi/2} \frac{x^2\cos x}{1+e^x}\,dx.$ Now under $x\mapsto -x$,

• $x^2$ remains $x^2$,
• $\cos(-x)=\cos x$,
• $e^{-x}$ appears in the denominator.

So, $I=\int_{-\pi/2}^{\pi/2} \frac{x^2\cos x}{1+e^{-x}}\,dx.$

3. Add the two expressions for $I$: $2I=\int_{-\pi/2}^{\pi/2} x^2\cos x\left(\frac{1}{1+e^x}+\frac{1}{1+e^{-x}}\right)dx.$

Now simplify the bracket: $\frac{1}{1+e^x}+\frac{1}{1+e^{-x}}=\frac{1}{1+e^x}+\frac{e^x}{1+e^x}=1.$

Hence, $2I=\int_{-\pi/2}^{\pi/2} x^2\cos x\,dx.$ Therefore, $I=\frac12\int_{-\pi/2}^{\pi/2} x^2\cos x\,dx.$

4. Since $x^2\cos x$ is an even function, $I=\frac12\cdot 2\int_0^{\pi/2} x^2\cos x\,dx=\int_0^{\pi/2} x^2\cos x\,dx.$

So we now evaluate $\int_0^{\pi/2} x^2\cos x\,dx.$

5. Use integration by parts. Let $u=x^2,\quad dv=\cos x\,dx.$ Then $du=2x\,dx,\quad v=\sin x.$ Thus, $\int x^2\cos x\,dx=x^2\sin x-\int 2x\sin x\,dx.$

Again apply integration by parts to $\int 2x\sin x\,dx$: Let $u=2x,\quad dv=\sin x\,dx,$ so $du=2\,dx,\quad v=-\cos x.$ Hence, $\int 2x\sin x\,dx=-2x\cos x+2\int \cos x\,dx=-2x\cos x+2\sin x.$

Therefore, $\int x^2\cos x\,dx=x^2\sin x+2x\cos x-2\sin x.$

6. Evaluate from $0$ to $\pi/2$: $\int_0^{\pi/2} x^2\cos x\,dx=\left[x^2\sin x+2x\cos x-2\sin x\right]_0^{\pi/2}.$ At $x=\pi/2$, $\sin\frac\pi2=1,\quad \cos\frac\pi2=0,$ so value is $\frac{\pi^2}{4}(1)+2\cdot \frac\pi2\cdot 0-2(1)=\frac{\pi^2}{4}-2.$ At $x=0$, $0^2\sin0+2\cdot0\cos0-2\sin0=0.$

Thus, $I=\frac{\pi^2}{4}-2.$

7. Compare with options:

• A: $\dfrac{\pi^2}{4}-2$ ✔
• B: $\dfrac{\pi^2}{4}+2$ ✘
• C: $\pi^2-e^{\pi/2}$ ✘
• D: $\pi^2+e^{\pi/2}$ ✘

So the correct option is A.