Question

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

Answer 1

First assume the given integral as I. Now take $x=-t$ substitution. Find $dx$ , limits in terms of t. Substitute them all. Now try to find a relation between this new integral and I. By using this relation try to eliminate the exponential term. Now you have I as an integral of normal constants. Use basic integration properties to solve the I. Find the value of I (after integrating) by using general algebra. Use the following integral:
$\int{kdx=kx+c}$

Complete step by step answer:
Given integral in the question, which we need to solve:
$\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\dfrac{1}{{{e}^{\sin x}}+1}dx}$.
Assume this integral as I, now I can be written as:
$I=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\dfrac{1}{{{e}^{\sin x}}+1}dx}$.………………………………..(i)
Now assume $x=-t$ . By differentiating this we get $dx=-dt$ .
Now limits $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ will be multiplied by $''-1''$ times that is $\left( \dfrac{\pi }{2},-\dfrac{\pi }{2} \right)$.
By substituting the above values into integration, we get:
$\Rightarrow \int\limits_{\dfrac{\pi }{2}}^{-\dfrac{\pi }{2}}{\dfrac{-dt}{{{e}^{-\sin t}}+1}}$
By basic integration property, we can say the following:
$\Rightarrow \int\limits_{a}^{b}{-f\left( x \right)}dx=\int\limits_{b}^{a}{f\left( x \right)}dx$
By using this our integral becomes as it is given below:
$\Rightarrow I=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\dfrac{dt}{{{e}^{-\sin t}}+1}}$
By taking least common multiple in the denominator, we get:
$\Rightarrow I=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\dfrac{{{e}^{\sin t}}dt}{{{e}^{\sin t}}+1}}$
By substituting t as x using the dummy rule in above integral, we can write it as:
$\Rightarrow I=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\dfrac{{{e}^{\sin x}}dx}{{{e}^{\sin x}}+1}}$……….(ii)
Now, by adding equation (i) and equation (ii), we get it as
$\Rightarrow I+I=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\dfrac{dx}{{{e}^{\sin x}}+1}}+\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\dfrac{{{e}^{\sin x}}dx}{{{e}^{\sin x}}+1}}$
By simplifying the above equation, we get it as:
$\Rightarrow 2I=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\dfrac{{{e}^{\sin x}}+1}{{{e}^{\sin x}}+1}dx}$
By cancelling common terms in right hand side, we get:
$\Rightarrow 2I=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{dx}$
By basic knowledge of integration, we can say the formula:
$\Rightarrow \int{kdx=kx+c}$
By using the above formula, we can write equation:
$\Rightarrow 2I=\left[ x \right]_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}$
By substituting the limits, we can write it as:
$\Rightarrow 2I=\dfrac{\pi }{2}-\left( -\dfrac{\pi }{2} \right)=\pi$
By dividing with 2 on both sides of equation, we get:
$\dfrac{2I}{2}=\dfrac{\pi }{2}$
By simplifying it, finally we can say value of I to be:
$\therefore I=\dfrac{\pi }{2}$
Therefore, the value of the given integral is $\dfrac{\pi }{2}$ .
Option (d) is the correct answer for a given question.

Note:
The idea of seeing that if we substitute $-x$ the limits remain the same and we can cancel the exponential is very important. These kinds of tricks are very useful in integration. These tricks can be obtained only by lots of practice. Be careful while adding you must add LHS also generally students write I instead of 2I. Take the limits properly as the whole result depends on them.