
Find the value of \[\int_0^\pi {x\log \sin x} dx\].
A. \[\dfrac{\pi }{2}\log \dfrac{1}{2}\]
B. \[ - \dfrac{{{\pi ^2}}}{2}\log 2\]
C. \[\pi \log \dfrac{1}{2}\]
D. \[{\pi ^2}\log \dfrac{1}{2}\]
Answer
164.1k+ views
Hint: In this question, we have \[\int_0^\pi {x\log \sin x} dx\]. So, we will consider this equation as 1. Now we will use the identity \[\int_0^a {f\left( x \right)dx = } \int_0^a {f\left( {a - x} \right)dx} \]. After that, we will get equation 2.
Now we will add equations 1 and 2 and then we will use the identity \[\int\limits_0^{2a} {f\left( x \right)dx = 2\int\limits_0^a {f\left( x \right)dx} } \]. At last, we will obtain the final result.
Formula Used:We will use the following formulas:
1)\[\int_0^a {f\left( x \right)dx = } \int_0^a {f\left( {a - x} \right)dx} \]
2) \[\sin \left( {\pi - x} \right) = \sin x\]
3) \[\int\limits_0^{2a} {f\left( x \right)dx = 2\int\limits_0^a {f\left( x \right)dx} } \]
Complete step by step solution:Let us consider \[I = \int_0^\pi {x\log \left( {\sin x} \right)} dx\] ….... (1)
First, we will use the identity \[\int_0^a {f\left( x \right)dx = } \int_0^a {f\left( {a - x} \right)dx} \].
\[I = \int_0^\pi {\left( {\pi - x} \right)\log \left( {\sin \left( {\pi - x} \right)} \right)} dx\]
As we know that \[\sin \left( {\pi - x} \right) = \sin x\].
So, we get
\[I = \int_0^\pi {\left( {\pi - x} \right)\log \left( {\sin x} \right)} dx\] ….... (2)
Now, we will add equations (1) and (2).
\[ \Rightarrow 2I = \int_0^\pi {\left( {x + \pi - x} \right)\log \left( {\sin x} \right)} dx\]
\[ \Rightarrow 2I = \pi \int_0^\pi {\log \left( {\sin x} \right)} dx\]
Further, we will use the identity \[\int\limits_0^{2a} {f\left( x \right)dx = 2\int\limits_0^a {f\left( x \right)dx} } \] to get
\[ \Rightarrow 2I = \pi \cdot 2\int_0^{\dfrac{\pi }{2}} {\log \left( {\sin x} \right)} dx\]
\[ \Rightarrow 2I = 2\pi \int_0^{\dfrac{\pi }{2}} {\log \left( {\sin x} \right)} dx\]
We will further integrate 1 with respect to x.
\[ \Rightarrow I = \pi \left( { - \dfrac{\pi }{2}\log 2} \right)\]
\[ \Rightarrow I = - \dfrac{{{\pi ^2}}}{2}\log \left( 2 \right)\]
Option ‘C’ is correct
Note: As the functions change, it is not essential to integrate taking limits directly \[0\] to \[\dfrac{\pi }{2}\]. We must avoid errors and confusion while dealing with sign changes that occurred during integration. We need to adopt the identities that make the problems easier. Students are advised to write the correct identity and then solve the problem. Don’t choose the identity that makes the question more complicated.
Now we will add equations 1 and 2 and then we will use the identity \[\int\limits_0^{2a} {f\left( x \right)dx = 2\int\limits_0^a {f\left( x \right)dx} } \]. At last, we will obtain the final result.
Formula Used:We will use the following formulas:
1)\[\int_0^a {f\left( x \right)dx = } \int_0^a {f\left( {a - x} \right)dx} \]
2) \[\sin \left( {\pi - x} \right) = \sin x\]
3) \[\int\limits_0^{2a} {f\left( x \right)dx = 2\int\limits_0^a {f\left( x \right)dx} } \]
Complete step by step solution:Let us consider \[I = \int_0^\pi {x\log \left( {\sin x} \right)} dx\] ….... (1)
First, we will use the identity \[\int_0^a {f\left( x \right)dx = } \int_0^a {f\left( {a - x} \right)dx} \].
\[I = \int_0^\pi {\left( {\pi - x} \right)\log \left( {\sin \left( {\pi - x} \right)} \right)} dx\]
As we know that \[\sin \left( {\pi - x} \right) = \sin x\].
So, we get
\[I = \int_0^\pi {\left( {\pi - x} \right)\log \left( {\sin x} \right)} dx\] ….... (2)
Now, we will add equations (1) and (2).
\[ \Rightarrow 2I = \int_0^\pi {\left( {x + \pi - x} \right)\log \left( {\sin x} \right)} dx\]
\[ \Rightarrow 2I = \pi \int_0^\pi {\log \left( {\sin x} \right)} dx\]
Further, we will use the identity \[\int\limits_0^{2a} {f\left( x \right)dx = 2\int\limits_0^a {f\left( x \right)dx} } \] to get
\[ \Rightarrow 2I = \pi \cdot 2\int_0^{\dfrac{\pi }{2}} {\log \left( {\sin x} \right)} dx\]
\[ \Rightarrow 2I = 2\pi \int_0^{\dfrac{\pi }{2}} {\log \left( {\sin x} \right)} dx\]
We will further integrate 1 with respect to x.
\[ \Rightarrow I = \pi \left( { - \dfrac{\pi }{2}\log 2} \right)\]
\[ \Rightarrow I = - \dfrac{{{\pi ^2}}}{2}\log \left( 2 \right)\]
Option ‘C’ is correct
Note: As the functions change, it is not essential to integrate taking limits directly \[0\] to \[\dfrac{\pi }{2}\]. We must avoid errors and confusion while dealing with sign changes that occurred during integration. We need to adopt the identities that make the problems easier. Students are advised to write the correct identity and then solve the problem. Don’t choose the identity that makes the question more complicated.
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 Main 2025 Session 2: Exam Date, Admit Card, Syllabus, & More

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

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

Trending doubts
Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

JEE Main Chemistry Question Paper with Answer Keys and Solutions

JEE Main Reservation Criteria 2025: SC, ST, EWS, and PwD Candidates

JEE Mains 2025 Cut-Off GFIT: Check All Rounds Cutoff Ranks

Lami's Theorem

Other Pages
Total MBBS Seats in India 2025: Government College Seat Matrix

NEET Total Marks 2025: Important Information and Key Updates

Neet Cut Off 2025 for MBBS in Tamilnadu: AIQ & State Quota Analysis

Karnataka NEET Cut off 2025 - Category Wise Cut Off Marks

NEET Marks vs Rank 2024|How to Calculate?

NEET 2025: All Major Changes in Application Process, Pattern and More
