
What is the value of the integral \[\int\limits_0^\pi {x\sin xdx} \]?
A. \[\pi \]
B. 0
C. 1
D. \[{\pi ^2}\]
Answer
163.8k+ views
Hint: Here, a definite integral is given. First, apply the definite integration rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx\] and simplify the integral. After that, add this integral with the original integral and solve the integral. In the end, apply the upper and lower limits to get the required answer.
Formula Used: Definite integral rule: \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx\]
\[\int {\sin xdx = - \cos x} \]
Complete step by step solution: The given definite integral is \[\int\limits_0^\pi {x\sin xdx} \].
Let consider,
\[I = \int\limits_0^\pi {x\sin xdx} \] \[.....\left( 1 \right)\]
Now apply the definite integral rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx\] on the above integral.
\[I = \int\limits_0^\pi {\left( {\pi - x} \right)\sin \left( {\pi - x} \right)dx} \]
\[ \Rightarrow I = \int\limits_0^\pi {\left( {\pi - x} \right)\sin xdx} \] \[.....\left( 2 \right)\]
Add the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[2I = \int\limits_0^\pi {x\sin xdx} + \int\limits_0^\pi {\left( {\pi - x} \right)\sin xdx} \]
\[ \Rightarrow 2I = \int\limits_0^\pi {\left( {x + \pi - x} \right)\sin xdx} \]
\[ \Rightarrow 2I = \int\limits_0^\pi {\pi \sin xdx} \]
\[ \Rightarrow I = \dfrac{\pi }{2}\int\limits_0^\pi {\sin xdx} \]
Solve the integral.
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ { - \cos x} \right]_0^\pi \]
Apply the upper and lower limit.
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ { - \cos \pi + \cos 0} \right]\]
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ { - \left( { - 1} \right) + 1} \right]\]
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ 2 \right]\]
\[ \Rightarrow I = \pi \]
Therefore, \[\int\limits_0^\pi {x\sin xdx} = \pi \]
Option ‘A’ is correct
Note: students often get confused and solve the integral \[\int {\sin xdx = \cos x} \]. This formula is incorrect. Sometimes they forget to add the negative sign. The correct formula is \[\int {\sin xdx = - \cos x} \].
Formula Used: Definite integral rule: \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx\]
\[\int {\sin xdx = - \cos x} \]
Complete step by step solution: The given definite integral is \[\int\limits_0^\pi {x\sin xdx} \].
Let consider,
\[I = \int\limits_0^\pi {x\sin xdx} \] \[.....\left( 1 \right)\]
Now apply the definite integral rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx\] on the above integral.
\[I = \int\limits_0^\pi {\left( {\pi - x} \right)\sin \left( {\pi - x} \right)dx} \]
\[ \Rightarrow I = \int\limits_0^\pi {\left( {\pi - x} \right)\sin xdx} \] \[.....\left( 2 \right)\]
Add the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[2I = \int\limits_0^\pi {x\sin xdx} + \int\limits_0^\pi {\left( {\pi - x} \right)\sin xdx} \]
\[ \Rightarrow 2I = \int\limits_0^\pi {\left( {x + \pi - x} \right)\sin xdx} \]
\[ \Rightarrow 2I = \int\limits_0^\pi {\pi \sin xdx} \]
\[ \Rightarrow I = \dfrac{\pi }{2}\int\limits_0^\pi {\sin xdx} \]
Solve the integral.
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ { - \cos x} \right]_0^\pi \]
Apply the upper and lower limit.
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ { - \cos \pi + \cos 0} \right]\]
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ { - \left( { - 1} \right) + 1} \right]\]
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ 2 \right]\]
\[ \Rightarrow I = \pi \]
Therefore, \[\int\limits_0^\pi {x\sin xdx} = \pi \]
Option ‘A’ is correct
Note: students often get confused and solve the integral \[\int {\sin xdx = \cos x} \]. This formula is incorrect. Sometimes they forget to add the negative sign. The correct formula is \[\int {\sin xdx = - \cos x} \].
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

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

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

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

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

Degree of Dissociation and Its Formula With Solved Example for JEE

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

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series
