The value of ∫-π2π2x2cosx1+exdx is equal to

I=∫−π2π2x2cosx1+exdx=∫−π2π2x2cosx1+e−x since ∫abfxdx=∫abfa+b-xdx2I=∫−π2π2x2cosxdx I=∫0π2x2cosxdxI=x2sinx0π2+∫0π22xsinxdx = π24+ 2x−cosx0π2 +2∫0π2−cosxdx=-2+π24

The value of ∫-π2π2x2cosx1+exdx is equal to

I=∫−π2π2x2cosx1+exdx=∫−π2π2x2cosx1+e−x since ∫abfxdx=∫abfa+b-xdx2I=∫−π2π2x2cosxdx I=∫0π2x2cosxdxI=x2sinx0π2+∫0π22xsinxdx = π24+ 2x−cosx0π2 +2∫0π2−cosxdx=-2+π24

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$The value of \int_{\frac{- π}{2}}^{\frac{π}{2}} \frac{x^{2} \cos x}{1 + e^{x}} d x is equal to$

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$\frac{π^{2}}{4} - 2$

$\frac{π^{2}}{4} + 2$

$π^{2} - e^{\frac{π}{2}}$

$π^{2} + e^{\frac{π}{2}}$

answer is A.

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Detailed Solution

$\begin{array}{l} I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{x^{2} \cos x}{1 + e^{x}} d x = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{x^{2} \cos x}{1 + e^{- x}} (\sin c e \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x) \\ 2 I = \int_{- \frac{π}{2}}^{\frac{π}{2}} x^{2} \cos x d x \\ I = \int_{0}^{\frac{π}{2}} x^{2} \cos x d x \\ I = {[x^{2} \sin x]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} 2 x \sin x d x \\ = \frac{π^{2}}{4} + {[2 x (- \cos x)]}_{0}^{\frac{π}{2}}_{} + 2 \int_{0}^{\frac{π}{2}} (- \cos x) d x = - 2 + \frac{π^{2}}{4} \end{array}$