
Let \[{I_1} = \int_a^{\pi - a} {xf\left( {\sin x} \right)dx} \], \[{I_2} = \int_a^{\pi - a} {f\left( {\sin x} \right)dx} \], then find the value of \[{I_2}\] in term of \[{I_1}\].
A. \[\dfrac{\pi }{2}{I_1}\]
B. \[\pi {I_1}\]
C. \[\dfrac{2}{\pi }{I_1}\]
D. \[2{I_1}\]
Answer
163.8k+ views
Hint: To solve the given question we will apply definite integral property. First we will apply the property \[\int_b^a {f\left( x \right)dx} = \int_b^a {f\left( {a + b - x} \right)dx} \] in \[{I_1}\] and simplify it by using trigonometry supplementary formula. Then substitute \[\int_a^{\pi - a} {xf\left( {\sin x} \right)dx} = {I_1}\] and \[\int_a^{\pi - a} {f\left( {\sin x} \right)dx} = {I_2}\] and calculate the value of \[{I_2}\] in term \[{I_1}\].
Formula Used:Definite integral property:
\[\int_b^a {f\left( x \right)dx} = \int_b^a {f\left( {a + b - x} \right)dx} \]
Trigonometry supplementary formula:
\[\sin \left( {\pi - \theta } \right) = \sin \theta \]
Complete step by step solution:Given integration is \[{I_1} = \int_a^{\pi - a} {xf\left( {\sin x} \right)dx} \]
Now applying the definite integral property \[\int_b^a {f\left( x \right)dx} = \int_b^a {f\left( {a + b - x} \right)dx} \]:
\[ \Rightarrow {I_1} = \int_a^{\pi - a} {\left( {\pi - a + a - x} \right)f\left( {\sin \left( {\pi - a + a - x} \right)} \right)dx} \]
Simplify the above equation:
\[ \Rightarrow {I_1} = \int_a^{\pi - a} {\left( {\pi - x} \right)f\left( {\sin \left( {\pi - x} \right)} \right)dx} \]
Now we will apply trigonometry supplementary formula \[\sin \left( {\pi - \theta } \right) = \sin \theta \]:
\[ \Rightarrow {I_1} = \int_a^{\pi - a} {\left( {\pi - x} \right)f\left( {\sin x} \right)dx} \]
Now break the integration:
\[ \Rightarrow {I_1} = \int_a^{\pi - a} {\pi f\left( {\sin x} \right)dx} - \int_a^{\pi - a} {xf\left( {\sin x} \right)dx} \]
Rewrite the above equation:
\[ \Rightarrow {I_1} = \pi \int_a^{\pi - a} {f\left( {\sin x} \right)dx} - \int_a^{\pi - a} {xf\left( {\sin x} \right)dx} \]
Substitute \[\int_a^{\pi - a} {xf\left( {\sin x} \right)dx} = {I_1}\] and \[\int_a^{\pi - a} {f\left( {\sin x} \right)dx} = {I_2}\] in the above equation:
\[ \Rightarrow {I_1} = \pi {I_2} - {I_1}\]
Add \[{I_1}\] on both sides of the equation:
\[ \Rightarrow {I_1} + {I_1} = \pi {I_2} - {I_1} + {I_1}\]
\[ \Rightarrow 2{I_1} = \pi {I_2}\]
Divide both sides by \[\pi \]:
\[ \Rightarrow \dfrac{2}{\pi }{I_1} = {I_2}\]
Option ‘C’ is correct
Note: Students often mistake we they apply trigonometry supplementary formula. They confused with complementary formula. They applied \[\sin \left( {\pi - \theta } \right) = - \cos \theta \] which incorrect formula. The correct formula is \[\sin \left( {\pi - \theta } \right) = \sin \theta \].
Formula Used:Definite integral property:
\[\int_b^a {f\left( x \right)dx} = \int_b^a {f\left( {a + b - x} \right)dx} \]
Trigonometry supplementary formula:
\[\sin \left( {\pi - \theta } \right) = \sin \theta \]
Complete step by step solution:Given integration is \[{I_1} = \int_a^{\pi - a} {xf\left( {\sin x} \right)dx} \]
Now applying the definite integral property \[\int_b^a {f\left( x \right)dx} = \int_b^a {f\left( {a + b - x} \right)dx} \]:
\[ \Rightarrow {I_1} = \int_a^{\pi - a} {\left( {\pi - a + a - x} \right)f\left( {\sin \left( {\pi - a + a - x} \right)} \right)dx} \]
Simplify the above equation:
\[ \Rightarrow {I_1} = \int_a^{\pi - a} {\left( {\pi - x} \right)f\left( {\sin \left( {\pi - x} \right)} \right)dx} \]
Now we will apply trigonometry supplementary formula \[\sin \left( {\pi - \theta } \right) = \sin \theta \]:
\[ \Rightarrow {I_1} = \int_a^{\pi - a} {\left( {\pi - x} \right)f\left( {\sin x} \right)dx} \]
Now break the integration:
\[ \Rightarrow {I_1} = \int_a^{\pi - a} {\pi f\left( {\sin x} \right)dx} - \int_a^{\pi - a} {xf\left( {\sin x} \right)dx} \]
Rewrite the above equation:
\[ \Rightarrow {I_1} = \pi \int_a^{\pi - a} {f\left( {\sin x} \right)dx} - \int_a^{\pi - a} {xf\left( {\sin x} \right)dx} \]
Substitute \[\int_a^{\pi - a} {xf\left( {\sin x} \right)dx} = {I_1}\] and \[\int_a^{\pi - a} {f\left( {\sin x} \right)dx} = {I_2}\] in the above equation:
\[ \Rightarrow {I_1} = \pi {I_2} - {I_1}\]
Add \[{I_1}\] on both sides of the equation:
\[ \Rightarrow {I_1} + {I_1} = \pi {I_2} - {I_1} + {I_1}\]
\[ \Rightarrow 2{I_1} = \pi {I_2}\]
Divide both sides by \[\pi \]:
\[ \Rightarrow \dfrac{2}{\pi }{I_1} = {I_2}\]
Option ‘C’ is correct
Note: Students often mistake we they apply trigonometry supplementary formula. They confused with complementary formula. They applied \[\sin \left( {\pi - \theta } \right) = - \cos \theta \] which incorrect formula. The correct formula is \[\sin \left( {\pi - \theta } \right) = \sin \theta \].
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

Instantaneous Velocity - Formula based Examples for JEE

JEE Advanced 2025 Notes

JEE Main Chemistry Question Paper with Answer Keys and Solutions

Total MBBS Seats in India 2025: Government and Private Medical Colleges
