
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
162k+ views
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.
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.
Recently Updated Pages
If tan 1y tan 1x + tan 1left frac2x1 x2 right where x frac1sqrt 3 Then the value of y is

Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

NIT Cutoff Percentile for 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

JEE Advanced 2025 Notes
