First slide

Evaluation of definite integrals

Question

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

Moderate

A

$2 π$
B

1
C

2
D

$\frac{π}{2}$

Solution

$\begin{matrix} I = \int_{- π / 2}^{π / 2} \frac{x \sin x}{e^{x} + 1} d x = \int_{- π / 2}^{π / 2} \frac{(- x) \sin (- x)}{e^{- x} + 1} d x (Prop. 11) \\ = \int_{- π / 2}^{π / 2} \frac{(x \sin x) e^{x}}{e^{x} + 1} d x \end{matrix}$
$\begin{array}{l} 2 I = \int_{- π / 2}^{π / 2} \frac{x \sin x}{e^{x} + 1} + \int_{- π / 2}^{π / 2} \frac{e^{x} (x \sin x)}{e^{x} + 1} d x \\ = \int_{- π / 2}^{π / 2} \frac{(e^{x} + 1) (x \sin x)}{e^{x} + 1} d x \\ = \int_{- π / 2}^{π / 2} x \sin x d x = 2 \int_{0}^{π / 2} x \sin x d x \\ (Prop. 12) \\ = 2 [- {x \cos x|}_{0}^{π / 2} + \int_{0}^{π / 2} \cos x d x] \\ = {2 \sin x|}_{0}^{π / 2} = 2 \\ \Rightarrow I = 1 . \end{array}$

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