Question

What is the value of the integration \[\int\limits_0^{2\pi } {\left[ {\dfrac{{dx}}{{\left( {{e^{\sin x}} + 1} \right)}}} \right]} \] ?
A. \[\pi \]
B. 0
C. \[2\pi \]
D. \[\dfrac{\pi }{2}\]

Answer 1

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

Formula used:
\[\int\limits_0^a {f\left( x \right)dx} = \int\limits_0^a {f\left( {a - x} \right)dx} \]
\[\sin\left( {2\pi - x} \right) = - \sin x\]
\[\int\limits_a^b {dx} = \left[ x \right]_a^b\]

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

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

Note: Students often get confused about the trigonometric identity of \[\sin\left( {2\pi - x} \right)\] that whether \[\sin x\] or \[ - \sin x\]. But the correct identity is \[\sin\left( {2\pi - x} \right) = - \sin x\].
This identity is derived from the trigonometric property \[\sin \left( {A - B} \right) = \sin A \cos B - \cos A \sin B\].