The value of $I=I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^{-x}} \mathrm{~d} x$ is equal to

The value of $I=I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^{-x}} \mathrm{~d} x$ is equal to
  1. $\frac{\pi^2}{4}-2$
  2. $\frac{\pi^2}{4}+2$
  3. $\pi^2-\mathrm{e}^{\frac{\pi}{2}}$
  4. $\pi^2+\mathrm{e}^{\frac{\pi}{2}}$

Solution

$\begin{array}{r} I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^{-x}} \mathrm{~d} x ...(i)\\ I=\frac{\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\left(\frac{-\pi}{2}+\frac{\pi}{2}-x\right)^2 \cdot \cos \left(-\frac{\pi}{2}+\frac{\pi}{2}-x\right)}{1+e^{\left(-\frac{\pi}{2}+\frac{\pi}{2}+x\right)}}}{\quad \ldots\left[\int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right]} \end{array}$ $=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+\mathrm{e}^x}$ Adding equation (i) and (ii), we get $\begin{aligned} & 2 \mathrm{I}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(\frac{x^2 \cos x}{1+\mathrm{e}^x}+\frac{x^2 \cos x}{1+\mathrm{e}^{-x}}\right) \mathrm{d} x \\ & 2 \mathrm{I}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x\left[\frac{1}{1+\mathrm{e}^x}+\frac{1}{1+\mathrm{e}^{-x}}\right] \mathrm{d} x\end{aligned}$ $\begin{aligned} & 2 \mathrm{I}=\int_{\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x\left[\frac{1}{1+\mathrm{e}^x}+\frac{\mathrm{e}^x}{\mathrm{e}^x+1}\right] \mathrm{d} x \\ \therefore \quad & 2 \mathrm{I}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cdot \cos x \mathrm{~d} x\end{aligned}$ $\therefore \quad 2 \mathrm{I}=2 \int_0^{\frac{\pi}{2}} x^2 \cos x \mathrm{~d} x$ $\therefore\left[\begin{array}{l}\int_{-\mathrm{a}}^{\mathrm{a}} \mathrm{f}(x)=2 \int_0^{\mathrm{a}} \mathrm{f}(x) \mathrm{d} x \\ \text { if } \mathrm{f}(x) \text { is even function }\end{array}\right]$ $\begin{aligned} \therefore \quad I & =\int_0^{\frac{\pi}{2}} x^2 \cos x \cdot d x \\ & =\left[x^2 \cdot \sin x-2 \int x \cdot \sin x \mathrm{~d} x\right]_0^{\frac{\pi}{2}} \\ & =\left[x^2 \sin x-2\left(-x \cos x+\int \cos x \mathrm{~d} x\right)\right]_0^{\frac{\pi}{2}} \\ & =\left[x^2 \sin x+2 x \cos x-2 \sin x\right]_0^{\frac{\pi}{2}} \\ & =\left(\frac{\pi^2}{4}-2 \sin \frac{\pi}{2}-0+0-0\right) \\ & =\frac{\pi^2}{4}-2\end{aligned}$

Asked in: MHT CET 2024 (09 May Shift 1)

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