The value of is equal to

Engineering

Mathematics

Properties of Definite Integral

Question

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

$\frac{π^{2}}{4} + 2$

$π^{2} - e^{\frac{π}{2}}$

$\frac{π^{2}}{4} - 2$

$π^{2} + e^{\frac{π}{2}}$

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I = $\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{x^{2} cosx}{1 + e^{x}} dx$

Using King

I = $\int_{\frac{- π}{2}}^{\frac{π}{2}} \frac{x^{2} cosx}{1 + e^{- x}} dx$

2I = $\int_{\frac{- π}{2}}^{\frac{π}{2}} x^{2} cosx dx$

2I = 2 $\int_{0}^{\frac{π}{2}} x^{2} cosx dx \Rightarrow I = \int_{0}^{\frac{π}{2}} x^{2} cosx dx$

I = x² sin x – $\int 2 x sinx dx$ = x² sin x – 2 [– x cos x + $\int cosx dx$ ]

$= {(x^{2} sinx + 2 xcosx - 2 sinx)}_{0}^{π / 2} = \frac{π^{2}}{4} + 0 - 2 - 0 = \frac{π^{2}}{4} - 2 .$