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Find the value of
\[\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{2}}x}{1+{{\left( 2017 \right)}^{x}}}}dx\]

Last updated date: 15th Jun 2024
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Hint: We start solving the problem by using the property \[\int\limits_{-a}^{a}{f(x)dx=}\int\limits_{0}^{a}{\left( f(x)+f(-x) \right)}dx\] in the given definite integral. We make subsequent calculations required and use the property \[{{\sin }^{2}}\theta =\dfrac{\left( 1-\cos 2\theta \right)}{2}\] to proceed further through the problem. We then use the property \[\int\limits_{b}^{a}{\left( mf(x)+ng(x) \right)dx=}\int\limits_{b}^{a}{mf(x)dx+\int\limits_{b}^{a}{ng(x)dx}}\] and integrate the functions. We then substitute the limits of the definite integral in the obtained functions to get the required answer.

Complete step-by-step answer:
As mentioned in the question, we have to evaluate the following expression \[I=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{2}}x}{1+{{\left( 2017 \right)}^{x}}}}dx\].
Let us assume the value of the given definite integral be I. So, we get \[I=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{2}}x}{1+{{\left( 2017 \right)}^{x}}}}dx\].
Using the property of definite integration in the above definite integral, we can write as the following expression
\[\Rightarrow I=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{2}}x}{1+{{\left( 2017 \right)}^{x}}}}dx\].
By using the following property of definite integration \[\int\limits_{-a}^{a}{f(x)dx=}\int\limits_{0}^{a}{\left( f(x)+f(-x) \right)}dx\].
\[\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{{{\sin }^{2}}x}{1+{{\left( 2017 \right)}^{x}}}+\dfrac{{{\sin }^{2}}\left( -x \right)}{1+{{\left( 2017 \right)}^{\left( -x \right)}}} \right)}dx\] ---(1).
We know that ${{a}^{-x}}=\dfrac{1}{{{a}^{x}}}$. We use this result in equation (1).
\[\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{{{\sin }^{2}}x}{1+{{\left( 2017 \right)}^{x}}}+\dfrac{{{\left( -\sin \left( x \right) \right)}^{2}}}{1+\dfrac{1}{{{\left( 2017 \right)}^{x}}}} \right)}dx\].
\[\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{{{\sin }^{2}}x}{1+{{\left( 2017 \right)}^{x}}}+\dfrac{{{\sin }^{2}}x}{\dfrac{{{\left( 2017 \right)}^{x}}+1}{{{\left( 2017 \right)}^{x}}}} \right)}dx\].
\[\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{{{\sin }^{2}}x}{1+{{\left( 2017 \right)}^{x}}}+\dfrac{\left( {{\sin }^{2}}x \right)\times {{\left( 2017 \right)}^{x}}}{1+{{\left( 2017 \right)}^{x}}} \right)}dx\].
We now take ${{\sin }^{2}}x$ common in both terms present in integrand.
\[\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( {{\sin }^{2}}x \right)\left( \dfrac{1}{1+{{\left( 2017 \right)}^{x}}}+\dfrac{{{\left( 2017 \right)}^{x}}}{1+{{\left( 2017 \right)}^{x}}} \right)}dx\].
\[\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( {{\sin }^{2}}x \right)\left( \dfrac{1+{{\left( 2017 \right)}^{x}}}{1+{{\left( 2017 \right)}^{x}}} \right)}dx\].
\[\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( {{\sin }^{2}}x \right)}dx\].
Now, we will use the relation between cos \[2\theta \] and \[{{\sin }^{2}}\theta \] as mentioned in the hint above as
\[{{\sin }^{2}}\theta =\dfrac{\left( 1-\cos 2\theta \right)}{2}\].
 Hence, now we can simplify the definite integral as
\[\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{1-\cos 2x}{2} \right)}dx\].
Now, we use another property $\int\limits_{a}^{b}{\left( mf\left( x \right)+ng\left( x \right) \right)dx=\int\limits_{a}^{b}{mf\left( x \right)dx+\int\limits_{a}^{b}{ng\left( x \right)dx}}}$.
\[\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{1}{2} \right)}dx-\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{\cos 2x}{2} \right)}dx\ \ \ \ ...(a)\]
We know that $\int{adx=ax+C}$ and $\int{\cos axdx=\dfrac{\sin ax}{a}+C}$. We also know that $\int\limits_{a}^{b}{{{f}^{'}}\left( x \right)dx=\left[ f\left( x \right) \right]_{a}^{b}=f\left( b \right)-f\left( a \right)}$. We use this results in equation (a).
\[\Rightarrow I=\dfrac{1}{2}\left[ x \right]_{0}^{\dfrac{\pi }{2}}-\dfrac{1}{4}\left[ \sin 2x \right]_{0}^{\dfrac{\pi }{2}}\].
\[\Rightarrow I=\dfrac{1}{2}\left[ \dfrac{\pi }{2}-0 \right]-\dfrac{1}{4}\left[ \sin \pi -\sin 0 \right]\].
\[\Rightarrow I=\dfrac{1}{2}\left[ \dfrac{\pi }{2} \right]-\dfrac{1}{4}\left[ 0-0 \right]\].
\[\Rightarrow I=\dfrac{\pi }{4}-0\].
\[\Rightarrow I=\dfrac{\pi }{4}\].
Hence, the solution of the definite integral is \[\dfrac{\pi }{4}\].

Note: We should not make an error while using the properties mentioned in the solution. The property of definite integration that would be useful in solving this question would be
\[\int\limits_{-a}^{a}{f(x)dx=}\int\limits_{0}^{a}{\left( f(x)+f(-x) \right)}dx\]. This property is usually used when the upper limit and lower limit have the same absolute values with opposite signs. Without the use of these properties the solution would entangle and the student might get a wrong solution.