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Integration of Piecewise Continuous Functions

The value of ...

Question

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

A

0

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B

1

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C

- \frac{π}{2}

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D

\frac{π}{2}

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Solution

The correct option is B $\frac{π}{2}$
Let $I = \int_{\frac{- π}{2}}^{\frac{π}{2}} \frac{1}{e^{sin x} + 1} d x$ ..... $(1)$
$\Rightarrow I = \int_{\frac{- π}{2}}^{\frac{π}{2}} \frac{1}{e^{sin (\frac{π}{2} - \frac{π}{2} - x)} + 1} d x$
$\Rightarrow= \int_{\frac{- π}{2}}^{\frac{π}{2}} \frac{e^{sin x}}{e^{sin x} + 1} d x$ ..... $(2)$
Adding eqns $(1)$ and $(2)$ we get
$2 I = \int_{\frac{- π}{2}}^{\frac{π}{2}} \frac{1}{e^{sin x} + 1} d x + \int_{\frac{- π}{2}}^{\frac{π}{2}} \frac{e^{sin x}}{e^{sin x} + 1} d x$
$2 I = \int_{\frac{- π}{2}}^{1 + \frac{π}{2}} \int_{\frac{- π}{2}}^{\frac{π}{2}} \frac{1 + e^{sin x}}{e^{sin x} + 1} d x$
$2 I = \int_{\frac{- π}{2}}^{\frac{π}{2}} 1. d x$
$2 I = {[x]}_{\frac{- π}{2}}^{\frac{π}{2}}$
$2 I = \frac{π}{2} + \frac{π}{2} = \frac{2 π}{2}$
$∴ I = \frac{π}{2}$

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