Question

What is the value of the integration \[\int\limits_{\dfrac{{ - \pi }}{2}}^{\dfrac{\pi }{2}} {\dfrac{1}{{\left( {1 + {e^{\sin x}}} \right)}}} dx\] ?
A. \[\dfrac{\pi }{2}\]
B. \[\dfrac{\pi }{4}\]
C. \[\pi \]
D. \[\dfrac{{3\pi }}{2}\]

Answer 1

Hint: First, apply the integration property \[\int\limits_{ - a}^a {f\left( x \right)dx} = \int\limits_0^a {\left[ {f\left( x \right) + f\left( { - x} \right)} \right]dx} \] and simplify the given integral. Then use the trigonometric identity of angle \[\sin\left( { - x} \right) = - \sin x\] and simplify the integral equation. After that, add both terms and solve the integral equation. In the end, integrate the given integral with respect to \[x\] and get the value of the given integral \[\int\limits_{\dfrac{{ - \pi }}{2}}^{\dfrac{\pi }{2}} {\dfrac{1}{{\left( {1 + {e^{\sin x}}} \right)}}} dx\].

Formula used:
\[\int\limits_{ - a}^a {f\left( x \right)dx} = \int\limits_0^a {\left[ {f\left( x \right) + f\left( { - x} \right)} \right]dx} \]
\[\sin\left( { - x} \right) = - \sin x\]
\[\int\limits_a^b {dx} = \left[ x \right]_a^b\]

Complete step by step solution:
The given integral is \[\int\limits_{\dfrac{{ - \pi }}{2}}^{\dfrac{\pi }{2}} {\dfrac{1}{{\left( {1 + {e^{\sin x}}} \right)}}} dx\].
Let \[I\] be the value of the above integral.
Then,
 \[I = \int\limits_{\dfrac{{ - \pi }}{2}}^{\dfrac{\pi }{2}} {\dfrac{1}{{\left( {1 + {e^{\sin x}}} \right)}}} dx\]
Now apply the integration property \[\int\limits_{ - a}^a {f\left( x \right)dx} = \int\limits_0^a {\left[ {f\left( x \right) + f\left( { - x} \right)} \right]dx} \].
We get,
\[I = \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\dfrac{1}{{\left( {1 + {e^{\sin x}}} \right)}} + \dfrac{1}{{\left( {1 + {e^{\sin\left( { - x} \right)}}} \right)}}} \right]} dx\]
Now simplify the above integral using the trigonometric property \[\sin\left( { - x} \right) = - \sin x\].
\[I = \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\dfrac{1}{{\left( {1 + {e^{\sin x}}} \right)}} + \dfrac{1}{{\left( {1 + {e^{ - \sin x}}} \right)}}} \right]} dx\]
Rewrite the term \[{e^{ - \sin x}}\] using the property \[{a^{ - m}} = \dfrac{1}{{{a^m}}}\] .
\[I = \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\dfrac{1}{{\left( {1 + {e^{\sin x}}} \right)}} + \dfrac{1}{{\left( {1 + \dfrac{1}{{{e^{\sin x}}}}} \right)}}} \right]} dx\]
\[ \Rightarrow I = \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\dfrac{1}{{\left( {1 + {e^{\sin x}}} \right)}} + \dfrac{1}{{\left( {\dfrac{{{e^{\sin x}} + 1}}{{{e^{\sin x}}}}} \right)}}} \right]} dx\]
\[ \Rightarrow I = \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\dfrac{1}{{\left( {1 + {e^{\sin x}}} \right)}} + \dfrac{{{e^{\sin x}}}}{{\left( {{e^{\sin x}} + 1} \right)}}} \right]} dx\]

Since the denominator of both terms is same. So, add the terms present in the numerator of both terms.
\[I = \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\dfrac{{1 + {e^{\sin x}}}}{{\left( {1 + {e^{\sin x}}} \right)}}} \right]} dx\]
\[I = \int\limits_0^{\dfrac{\pi }{2}} {\left[ 1 \right]} dx\]
Now integrate the above integral with respect to the variable \[x\].
\[I = \left[ x \right]_0^{\dfrac{\pi }{2}}\]
Apply the limits on right-hand side.
\[I = \dfrac{\pi }{2} - 0\]
\[ \Rightarrow I = \dfrac{\pi }{2}\]
Thus, the value of the given integral is,
\[\int\limits_{\dfrac{{ - \pi }}{2}}^{\dfrac{\pi }{2}} {\dfrac{1}{{\left( {1 + {e^{\sin x}}} \right)}}} dx = \dfrac{\pi }{2}\]
Hence the correct option is A.

Note: Students often get confused about the trigonometric identities of the opposite angles. The opposite angle identities change the trigonometric functions of negative angles to the functions of positive angles.
The opposite angle identities of the basic trigonometric identities are:
\[\sin\left( { - x} \right) = - \sin x\]
\[\cos\left( { - x} \right) = \cos x\]
\[\tan\left( { - x} \right) = - \tan x\]