
What is the value of the integral \[\int\limits_0^{\dfrac{\pi }{2}} {\left\{ {x - \left[ {\sin x} \right]} \right\}dx} \]?
A. \[\dfrac{{{\pi ^2}}}{8}\]
B. \[\dfrac{{{\pi ^2}}}{8} - 1\]
C. \[\dfrac{{{\pi ^2}}}{8} - 2\]
D. None of these
Answer
161.1k+ views
Hint: Here, a definite integral is given. First, simplify the given integral by using the subtraction rule of integration. Then, solve both integrals by using the standard formulas. In the end, substitute the upper and the lower limit values in the equation and get the required answer.
Formula Used: \[\int\limits_a^b {\left[ {f\left( x \right) \pm g\left( x \right)} \right]dx} = \int\limits_a^b {f\left( x \right)dx \pm } \int\limits_a^b {g\left( x \right)dx} \]
\[\int\limits_a^b {{x^n}dx} = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]_a^b\]
\[\int\limits_a^b {\sin xdx} = \left[ { - \cos x} \right]_a^b\]
Complete step by step solution: The given definite integral is \[\int\limits_0^{\dfrac{\pi }{2}} {\left\{ {x - \left[ {\sin x} \right]} \right\}dx} \].
Let’s simplify the given integral by using the integration rule.
\[\int\limits_0^{\dfrac{\pi }{2}} {\left\{ {x - \left[ {\sin x} \right]} \right\}dx} = \int\limits_0^{\dfrac{\pi }{2}} {xdx} - \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\sin x} \right]dx} \]
Solve the right-hand side.
Apply the power rule \[\int\limits_a^b {{x^n}dx} = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]_a^b\] for the first term and \[\int\limits_a^b {\sin xdx} = \left[ { - \cos x} \right]_a^b\] for the second term.
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{2}} {\left\{ {x - \left[ {\sin x} \right]} \right\}dx} = \left[ {\dfrac{{{x^{1 + 1}}}}{{1 + 1}}} \right]_0^{\dfrac{\pi }{2}} - \left[ { - \cos x} \right]_0^{\dfrac{\pi }{2}}\]
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{2}} {\left\{ {x - \left[ {\sin x} \right]} \right\}dx} = \left[ {\dfrac{{{x^2}}}{2}} \right]_0^{\dfrac{\pi }{2}} + \left[ {\cos x} \right]_0^{\dfrac{\pi }{2}}\]
Apply the upper and lower limits.
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{2}} {\left\{ {x - \left[ {\sin x} \right]} \right\}dx} = \left[ {\dfrac{{{{\left( {\dfrac{\pi }{2}} \right)}^2}}}{2} - \dfrac{{{{\left( 0 \right)}^2}}}{2}} \right] + \left[ {\cos \dfrac{\pi }{2} - \cos 0} \right]\]
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{2}} {\left\{ {x - \left[ {\sin x} \right]} \right\}dx} = \left[ {\dfrac{{{\pi ^2}}}{8} - 0} \right] + \left[ {0 - 1} \right]\]
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{2}} {\left\{ {x - \left[ {\sin x} \right]} \right\}dx} = \dfrac{{{\pi ^2}}}{8} - 1\]
Option ‘B’ is correct
Note: Students often get confused about the formula of the definite integral of the function. They used \[\int\limits_a^b {f\left( x \right)} dx = F\left( b \right) + F\left( a \right)\] , which is incorrect. The correct formula is \[\int\limits_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right)\].
Sometimes they also add integration constant \[c\] in the definite integral. But definite integral is calculated for a certain interval. So, there is no need to write the integration constant.
Formula Used: \[\int\limits_a^b {\left[ {f\left( x \right) \pm g\left( x \right)} \right]dx} = \int\limits_a^b {f\left( x \right)dx \pm } \int\limits_a^b {g\left( x \right)dx} \]
\[\int\limits_a^b {{x^n}dx} = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]_a^b\]
\[\int\limits_a^b {\sin xdx} = \left[ { - \cos x} \right]_a^b\]
Complete step by step solution: The given definite integral is \[\int\limits_0^{\dfrac{\pi }{2}} {\left\{ {x - \left[ {\sin x} \right]} \right\}dx} \].
Let’s simplify the given integral by using the integration rule.
\[\int\limits_0^{\dfrac{\pi }{2}} {\left\{ {x - \left[ {\sin x} \right]} \right\}dx} = \int\limits_0^{\dfrac{\pi }{2}} {xdx} - \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\sin x} \right]dx} \]
Solve the right-hand side.
Apply the power rule \[\int\limits_a^b {{x^n}dx} = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]_a^b\] for the first term and \[\int\limits_a^b {\sin xdx} = \left[ { - \cos x} \right]_a^b\] for the second term.
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{2}} {\left\{ {x - \left[ {\sin x} \right]} \right\}dx} = \left[ {\dfrac{{{x^{1 + 1}}}}{{1 + 1}}} \right]_0^{\dfrac{\pi }{2}} - \left[ { - \cos x} \right]_0^{\dfrac{\pi }{2}}\]
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{2}} {\left\{ {x - \left[ {\sin x} \right]} \right\}dx} = \left[ {\dfrac{{{x^2}}}{2}} \right]_0^{\dfrac{\pi }{2}} + \left[ {\cos x} \right]_0^{\dfrac{\pi }{2}}\]
Apply the upper and lower limits.
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{2}} {\left\{ {x - \left[ {\sin x} \right]} \right\}dx} = \left[ {\dfrac{{{{\left( {\dfrac{\pi }{2}} \right)}^2}}}{2} - \dfrac{{{{\left( 0 \right)}^2}}}{2}} \right] + \left[ {\cos \dfrac{\pi }{2} - \cos 0} \right]\]
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{2}} {\left\{ {x - \left[ {\sin x} \right]} \right\}dx} = \left[ {\dfrac{{{\pi ^2}}}{8} - 0} \right] + \left[ {0 - 1} \right]\]
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{2}} {\left\{ {x - \left[ {\sin x} \right]} \right\}dx} = \dfrac{{{\pi ^2}}}{8} - 1\]
Option ‘B’ is correct
Note: Students often get confused about the formula of the definite integral of the function. They used \[\int\limits_a^b {f\left( x \right)} dx = F\left( b \right) + F\left( a \right)\] , which is incorrect. The correct formula is \[\int\limits_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right)\].
Sometimes they also add integration constant \[c\] in the definite integral. But definite integral is calculated for a certain interval. So, there is no need to write the integration constant.
Recently Updated Pages
Crack JEE Advanced 2025 with Vedantu's Live Classes

JEE Advanced Maths Revision Notes

JEE Advanced Chemistry Revision Notes

The students S1 S2 S10 are to be divided into 3 groups class 11 maths JEE_Advanced

Solutions Class 12 Notes JEE Advanced Chemistry [PDF]

JEE Advanced 2022 Chemistry Question Paper 2 with Solutions

Trending doubts
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

IIT CSE Cutoff: Category‐Wise Opening and Closing Ranks

JEE Advanced Cut Off 2024

Other Pages
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

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