Indefinite Integration 1 Question 17
18. The value of $\int _{-\pi / 2}^{\pi / 2} \frac{x^{2} \cos x}{1+e^{x}} d x$ is equal to
(a) $\frac{\pi^{2}}{4}-2$
(b) $\frac{\Pi^{2}}{4}+2$
(c) $\pi^{2}-e^{-\pi / 2}$
(d) $\pi^{2}+e^{\pi / 2}$
(2016 Adv.)
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Solution:
- Let $I=\int _{-\pi / 2}^{\pi / 2} \frac{x^{2} \cos x}{1+e^{x}} d x$
$$ \begin{aligned} & \because \int _a^{b} f(x) d x=\int _a^{b} f(a+b-x) d x \\ & \Rightarrow I=\int _{-\pi / 2}^{\pi / 2} \frac{x^{2} \cos (-x)}{1+e^{-x}} d x \end{aligned} $$
On adding Eqs. (i) and (ii), we get
$$ \begin{aligned} & 2 I=\int _{-\pi / 2}^{\pi / 2} x^{2} \cos x \frac{1}{1+e^{x}}+\frac{1}{1+e^{-x}} d x \\ &=\int _{-\pi / 2}^{\pi / 2} x^{2} \cos x \cdot(1) d x \\ & \because \int _{-a}^{a} f(x) d x=2 \int _0^{a} f(x) d x, \text { when } f(-x)=f(x) \\ & \Rightarrow 2 I=2 \int _0^{\pi / 2} x^{2} \cos x d x \end{aligned} $$
Using integration by parts, we get
$$ \begin{aligned} 2 I & =2\left[x^{2}(\sin x)-(2 x)(-\cos x)+(2)(-\sin x)\right] _0^{\pi / 2} \\ \Rightarrow 2 I & =2 \frac{\pi^{2}}{4}-2 \\ \therefore \quad I & =\frac{\pi^{2}}{4}-2 \end{aligned} $$
