Question

The value of $\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{dx}}{{{e^{\sin x}} + 1}}}$ is equal to
(a) 0
(b) 1
(c) $- \dfrac{\pi }{2}$
(d) $\dfrac{\pi }{2}$

Answer 1

Here, we need to evaluate the given integral. Let. We will find the value of the negative value of the integral using the rules of exponents trigonometry. Then, we will use the properties of definite integrals to find the value of the given integral.

Formula Used: We will use the following formulas:
A) The sine of a negative angle $- \theta$ is equal to the negative of the sine of the positive angle $\theta$.
B) The number${a^{ - b}}$ is equal to the reciprocal of the number $a$ raised to the positive power $b$, that is ${a^{ - b}} = {\left( {\dfrac{1}{a}} \right)^b}$.
C) The integral of the form $\int\limits_{ - a}^a {f\left( x \right)} dx$ can be written as $\int\limits_{ - a}^a {f\left( x \right)} dx = \int\limits_0^a {\left[ {f\left( x \right) + f\left( { - x} \right)} \right]} dx$.

Complete step by step solution:
We will use the properties of definite integrals to simplify the given integral.
The given integral is $\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{dx}}{{{e^{\sin x}} + 1}}}$.
Let $f\left( x \right) = \dfrac{1}{{{e^{\sin x}} + 1}}$.
Substituting $x$ as $- x$, we get
$\Rightarrow f\left( { - x} \right) = \dfrac{1}{{{e^{\sin \left( { - x} \right)}} + 1}}$
The sine of a negative angle $- \theta$ is equal to the negative of the sine of the positive angle
$\theta$.
Thus, we get

$\sin \left( { - x} \right) = - \sin x$
Substituting $\sin \left( { - x} \right) = - \sin x$ in the equation $f\left( { - x} \right) = \dfrac{1}{{{e^{\sin \left( { - x} \right)}} + 1}}$, we get
$\Rightarrow f\left( { - x} \right) = \dfrac{1}{{{e^{ - \sin x}} + 1}}$
The number${a^{ - b}}$ is equal to the reciprocal of the number $a$ raised to the positive power $b$, that is ${a^{ - b}} = {\left( {\dfrac{1}{a}} \right)^b}$.
Therefore, we get
${e^{ - \sin x}} = \dfrac{1}{{{e^{\sin x}}}}$
Substituting ${e^{ - \sin x}} = \dfrac{1}{{{e^{\sin x}}}}$ in the equation $f\left( { - x} \right) = \dfrac{1}{{{e^{ - \sin x}} + 1}}$, we get
$\Rightarrow f\left( { - x} \right) = \dfrac{1}{{\dfrac{1}{{{e^{\sin x}}}} + 1}}$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow f\left( { - x} \right) = \dfrac{1}{{\dfrac{{1 + {e^{\sin x}}}}{{{e^{\sin x}}}}}}\\\ \Rightarrow f\left( { - x} \right) = \dfrac{{{e^{\sin x}}}}{{1 + {e^{\sin x}}}}\end{array}$
Now, we know that the integral of the form $\int\limits_{ - a}^a {f\left( x \right)} dx$ can be written as $\int\limits_{ - a}^a {f\left( x \right)} dx = \int\limits_0^a {\left[ {f\left( x \right) + f\left( { - x} \right)} \right]} dx$.
Therefore, substituting $a = \dfrac{\pi }{2}$, $f\left( x \right) = \dfrac{1}{{{e^{\sin x}} + 1}}$ and $f\left( { - x} \right) = \dfrac{{{e^{\sin x}}}}{{1 + {e^{\sin x}}}}$ in the formula, we get
$\Rightarrow \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{1}{{{e^{\sin x}} + 1}}} dx = \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\dfrac{1}{{{e^{\sin x}} + 1}} + \dfrac{{{e^{\sin x}}}}{{1 + {e^{\sin x}}}}} \right]} dx$
Adding the terms of the expression, we get
$\Rightarrow \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{1}{{{e^{\sin x}} + 1}}} dx = \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\dfrac{{1 + {e^{\sin x}}}}{{1 + {e^{\sin x}}}}} \right]} dx$
Simplifying the expression, we get
$\Rightarrow \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{1}{{{e^{\sin x}} + 1}}} dx = \int\limits_0^{\dfrac{\pi }{2}} {\left[ 1 \right]} dx$
The integral of a constant with respect to $x$ is $x$.
Therefore, we get
$\Rightarrow \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{1}{{{e^{\sin x}} + 1}}} dx = \left. {\left( x \right)} \right|_0^{\dfrac{\pi }{2}}$
Substituting the limits, we get
$\Rightarrow \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{1}{{{e^{\sin x}} + 1}}} dx = \dfrac{\pi }{2} - 0$
Therefore, we get
$\Rightarrow \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{1}{{{e^{\sin x}} + 1}}} dx = \dfrac{\pi }{2}$
Thus, the value of the integral $\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{dx}}{{{e^{\sin x}} + 1}}}$ is $\dfrac{\pi }{2}$.

$\therefore$ The correct option is option (d).

Note:
We used the property $\int\limits_{ - a}^a {f\left( x \right)} dx = \int\limits_0^a {\left[ {f\left( x \right) + f\left( { - x} \right)} \right]} dx$ to solve the given problem. This property is derived using the property of odd and even functions, that is $\int\limits_{ - a}^a {f\left( x \right)} dx = 2\int\limits_0^a {f\left( x \right)} dx$ if $f\left( x \right)$ is even, or $\int\limits_{ - a}^a {f\left( x \right)} dx = 0$ if $f\left( x \right)$ is odd. Since the graph of an even function is symmetric about the vertical axis, we get $\int\limits_0^a {f\left( x \right)} dx = \int\limits_{ - a}^0 {f\left( x \right)} dx$. Also, since the graph of an odd function is symmetric in opposite quadrants, we get $\int\limits_{ - a}^0 {f\left( x \right)} dx = - \int\limits_0^a {f\left( x \right)} dx$. From these equations, we get $\int\limits_{ - a}^a {f\left( x \right)} dx = \int\limits_0^a {\left[ {f\left( x \right) + f\left( { - x} \right)} \right]} dx$.