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

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.