
If \[\int\limits_0^\pi {xf\left( {\sin x} \right)} dx = A\int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)} dx\], then what is the value of \[A\]?
A. \[2\pi \]
B. \[\pi \]
C. \[\dfrac{\pi }{4}\]
D. 0
Answer
161.1k+ views
Hint: Here, an equation of the definite integral is given. First, simplify the left-hand side integral by applying the integration rule \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\]. Then, add the left-hand side and this simplified integral and solve it. After that, apply the integration rule \[\int\limits_0^{2a} {f\left( x \right)} dx = 2\int\limits_0^a {f\left( x \right)} dx\] if \[f\left( {2a - x} \right) = f\left( x \right)\] and solve the integral. In the end, compare the integral with the right-hand side of the originally given integral and get the required answer.
Formula Used:Integration rules:
\[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\]
\[\int\limits_0^{2a} {f\left( x \right)} dx = 2\int\limits_0^a {f\left( x \right)} dx\] if \[f\left( {2a - x} \right) = f\left( x \right)\]
Complete step by step solution:The given equation of the definite integral is \[\int\limits_0^\pi {xf\left( {\sin x} \right)} dx = A\int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)} dx\].
Let consider,
\[I = \int\limits_0^\pi {xf\left( {\sin x} \right)} dx\] \[.....\left( 1 \right)\]
Apply the integration rule \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\].
\[I = \int\limits_0^\pi {\left( {\pi - x} \right)f\left( {\sin \left( {\pi - x} \right)} \right)} dx\]
\[ \Rightarrow I = \int\limits_0^\pi {\left( {\pi - x} \right)f\left( {\sin x} \right)} dx\]
Now add the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[I + I = \int\limits_0^\pi {xf\left( {\sin x} \right)} dx + \int\limits_0^\pi {\left( {\pi - x} \right)f\left( {\sin x} \right)} dx\]
\[ \Rightarrow 2I = \int\limits_0^\pi {\left( {x + \pi - x} \right)f\left( {\sin x} \right)} dx\]
\[ \Rightarrow 2I = \int\limits_0^\pi {\pi f\left( {\sin x} \right)} dx\]
\[ \Rightarrow 2I = \pi \int\limits_0^\pi {f\left( {\sin x} \right)} dx\]
Here, \[f\left( {\sin x} \right) = f\left( {\sin \left( {\pi - x} \right)} \right)\] because \[\sin \left( {\pi - x} \right) = \sin x\] .
So, apply the integration rule \[\int\limits_0^{2a} {f\left( x \right)} dx = 2\int\limits_0^a {f\left( x \right)} dx\] if \[f\left( {2a - x} \right) = f\left( x \right)\].
We get,
\[ \Rightarrow 2I = 2\pi \int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)} dx\]
\[ \Rightarrow I = \pi \int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)} dx\]
Now compare the integral with the right-hand side of the originally given integral.
We get,
\[A\int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)} dx = \pi \int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)} dx\]
\[ \Rightarrow A = \pi \]
Option ‘A’ is correct
Note:
Below are some rules to find the definite integral of a function by splitting it into parts:
\[\int\limits_0^{2a} {f\left( x \right)} dx = \int\limits_0^a {f\left( x \right)} dx + \int\limits_0^a {f\left( {2a - x} \right)} dx\]
\[\int\limits_0^{2a} {f\left( x \right)} dx = 2\int\limits_0^a {f\left( x \right)} dx\] if \[f\left( {2a - x} \right) = f\left( x \right)\]
\[\int\limits_0^{2a} {f\left( x \right)} dx = 0\] if \[f\left( {2a - x} \right) = - f\left( x \right)\]
Formula Used:Integration rules:
\[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\]
\[\int\limits_0^{2a} {f\left( x \right)} dx = 2\int\limits_0^a {f\left( x \right)} dx\] if \[f\left( {2a - x} \right) = f\left( x \right)\]
Complete step by step solution:The given equation of the definite integral is \[\int\limits_0^\pi {xf\left( {\sin x} \right)} dx = A\int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)} dx\].
Let consider,
\[I = \int\limits_0^\pi {xf\left( {\sin x} \right)} dx\] \[.....\left( 1 \right)\]
Apply the integration rule \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\].
\[I = \int\limits_0^\pi {\left( {\pi - x} \right)f\left( {\sin \left( {\pi - x} \right)} \right)} dx\]
\[ \Rightarrow I = \int\limits_0^\pi {\left( {\pi - x} \right)f\left( {\sin x} \right)} dx\]
Now add the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[I + I = \int\limits_0^\pi {xf\left( {\sin x} \right)} dx + \int\limits_0^\pi {\left( {\pi - x} \right)f\left( {\sin x} \right)} dx\]
\[ \Rightarrow 2I = \int\limits_0^\pi {\left( {x + \pi - x} \right)f\left( {\sin x} \right)} dx\]
\[ \Rightarrow 2I = \int\limits_0^\pi {\pi f\left( {\sin x} \right)} dx\]
\[ \Rightarrow 2I = \pi \int\limits_0^\pi {f\left( {\sin x} \right)} dx\]
Here, \[f\left( {\sin x} \right) = f\left( {\sin \left( {\pi - x} \right)} \right)\] because \[\sin \left( {\pi - x} \right) = \sin x\] .
So, apply the integration rule \[\int\limits_0^{2a} {f\left( x \right)} dx = 2\int\limits_0^a {f\left( x \right)} dx\] if \[f\left( {2a - x} \right) = f\left( x \right)\].
We get,
\[ \Rightarrow 2I = 2\pi \int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)} dx\]
\[ \Rightarrow I = \pi \int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)} dx\]
Now compare the integral with the right-hand side of the originally given integral.
We get,
\[A\int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)} dx = \pi \int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)} dx\]
\[ \Rightarrow A = \pi \]
Option ‘A’ is correct
Note:
Below are some rules to find the definite integral of a function by splitting it into parts:
\[\int\limits_0^{2a} {f\left( x \right)} dx = \int\limits_0^a {f\left( x \right)} dx + \int\limits_0^a {f\left( {2a - x} \right)} dx\]
\[\int\limits_0^{2a} {f\left( x \right)} dx = 2\int\limits_0^a {f\left( x \right)} dx\] if \[f\left( {2a - x} \right) = f\left( x \right)\]
\[\int\limits_0^{2a} {f\left( x \right)} dx = 0\] if \[f\left( {2a - x} \right) = - f\left( x \right)\]
Recently Updated Pages
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 Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE 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

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

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

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

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

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

JEE Advanced 2025 Notes

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
