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

(C) $\pi^{2}-e^{\frac{\pi}{2}}$

(D) $\pi^{2}+e^{\frac{\pi}{2}}$

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Definite Integrals

The value of

\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{x ^{2} c o s x}{1 + e ^{x}} d x

is equal to

Answer: (A)

Use the property

\int_{- a}^{a} f (x) d x = \int_{0}^{a} [f (x) + f (- x)] d x

. Here

f (x) + f (- x) = \frac{x ^{2} c o s x}{1 + e ^{x}} + \frac{x ^{2} c o s x}{1 + e ^{- x}} = x^{2} cos x

. So the integral becomes

\int_{0}^{π /2} x^{2} cos x d x

. Using integration by parts twice:

\int x^{2} cos x d x = x^{2} sin x + 2 x cos x - 2 sin x + C

. Evaluating from 0 to

π /2

[\frac{π ^{2}}{4} \cdot 1 + 0 - 2 \cdot 1] - [0] = \frac{π ^{2}}{4} - 2

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Single correct option