Question

What is the value of the integral \[\int\limits_{ - 3}^3 {\dfrac{{{x^2}\sin 2x}}{{{x^2} + 1}}dx} \]?
A. 0
B. 1
C. \[2{\log _e}3\]
D. None of these

Answer 1

Hint: Here, a definite integral is given. First, check whether the function present in the given integral is odd or an even function. If the function is odd, then apply the property of the definite integral for the odd function. If the function is even, then apply the property of the definite integral for the even function and solve the integral to get the required answer.



Formula Used:\[\int\limits_{ - a}^a {f\left( x \right) dx} = 0\], if the function \[f\left( x \right)\] is an odd function. Means, \[f\left( { - x} \right) = - f\left( x \right)\]
\[\int\limits_{ - a}^a {f\left( x \right) dx} = 2\int\limits_0^a {f\left( x \right) dx} \], if the function \[f\left( x \right)\] is an even function. Means, \[f\left( { - x} \right) = f\left( x \right)\]



Complete step by step solution:The given definite integral is \[\int\limits_{ - 3}^3 {\dfrac{{{x^2}\sin 2x}}{{{x^2} + 1}}dx} \].

Let consider,
\[f\left( x \right) = \dfrac{{{x^2}\sin 2x}}{{{x^2} + 1}}\]
Now let’s calculate the value of \[f\left( { - x} \right)\].
\[f\left( { - x} \right) = \dfrac{{{{\left( { - x} \right)}^2}\sin \left( { - 2x} \right)}}{{{{\left( { - x} \right)}^2} + 1}}\]
\[ \Rightarrow f\left( { - x} \right) = \dfrac{{{x^2}\left( { - \sin 2x} \right)}}{{{x^2} + 1}}\]
\[ \Rightarrow f\left( { - x} \right) = - \dfrac{{{x^2}\sin 2x}}{{{x^2} + 1}}\]
\[ \Rightarrow f\left( { - x} \right) = - f\left( x \right)\]
Therefore, \[f\left( x \right) = \dfrac{{{x^2}\sin 2x}}{{{x^2} + 1}}\] is an odd function.
So, apply the integration formula \[\int\limits_{ - a}^a {f\left( x \right) dx} = 0\], if the function \[f\left( x \right)\] is an odd function.
We get,
\[\int\limits_{ - 3}^3 {\dfrac{{{x^2}\sin 2x}}{{{x^2} + 1}}dx} = 0\]



Option ‘A’ is correct



Note:Sometimes students get confused and try to solve the integral by using the trigonometric identities. We can solve this integral by using the methods of indefinite integral. But the answer will be wrong.
So, to calculate the correct answer in the definite integral, first check whether the given trigonometric function is odd or even.