Question

What is the value of \[\int_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{\sin x}}{{1 + {{\cos }^2}x}}{e^{ - {{\cos }^2}x}}dx} \]?
A. \[2{e^{ - 1}}\]
B. 1
C. 0
D. None of these

Answer 1

Hint: First we check whether the given function is an odd function or an even function by putting x = - x in the integrating term. Then we will apply the integration property to solve the given question.



Formula Used:Property of the definite integral:
 \[\int_{ - a}^a {f\left( x \right)dx} = \left\{ {\begin{array}{*{20}{c}}{0\,{\rm{if}}\,f\left( { - x} \right) = - f\left( x \right)}\\{2\int\limits_0^a {f\left( x \right)dx\,} {\rm{if}}\,f\left( { - x} \right) = f\left( x \right)}\end{array}} \right.\]



Complete step by step solution:Given definite integral is
\[\int_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{\sin x}}{{1 + {{\cos }^2}x}}{e^{ - {{\cos }^2}x}}dx} \]
Assume that, \[f\left( x \right) = \dfrac{{\sin x}}{{1 + {{\cos }^2}x}}{e^{ - {{\cos }^2}x}}\].
To check whether \[f\left( x \right)\] is an odd function or even function, we will put x = -x in \[f\left( x \right)\].
\[f\left( { - x} \right) = \dfrac{{\sin \left( { - x} \right)}}{{1 + {{\cos }^2}\left( { - x} \right)}}{e^{ - {{\cos }^2}\left( { - x} \right)}}\]
\[ \Rightarrow f\left( { - x} \right) = \dfrac{{ - \sin x}}{{1 + {{\cos }^2}x}}{e^{ - {{\cos }^2}x}}\]
Now putting \[\dfrac{{\sin x}}{{1 + {{\cos }^2}x}}{e^{ - {{\cos }^2}x}} = f\left( x \right)\]
\[ \Rightarrow f\left( { - x} \right) = - f\left( x \right)\]
The given function is an odd function.
Now applying the formula \[\int_{ - a}^a {f\left( x \right)dx} = \left\{ {\begin{array}{*{20}{c}}{0\,{\rm{if}}\,f\left( { - x} \right) = - f\left( x \right)}\\{2\int\limits_0^a {f\left( x \right)dx\,} {\rm{if}}\,f\left( { - x} \right) = f\left( x \right)}\end{array}} \right.\] in the given definite integral:
\[\int_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{\sin x}}{{1 + {{\cos }^2}x}}{e^{ - {{\cos }^2}x}}dx} = 0\]



Option ‘C’ is correct



Note: Students often make mistakes to solve the given definite integral. They used the substitution method by letting\[z = \cos x\]. But it is an incorrect way to solve it. We have to use the property of definite integral.