Question

Calculate the value of \[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\sin \left| x \right|dx} \].
A. 0
B. 1
C. 2
D. \[\pi \]

Answer 1

Hint First we will apply the absolute function in the \[\sin \left| x \right|\]. We know that, \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{ - x}&{x < 0}\\x&{x \ge 0}\end{array}} \right.\]. We will apply absolute function for \[\sin \left| x \right|\] in the interval \[\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]\].
According to absolute function, we will divide the integration into 2 integrations. Then will apply the formula \[\int {\sin x} = - \cos x + C\].

Formula used
Absolute function \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{ - x}&{x < 0}\\x&{x \ge 0}\end{array}} \right.\]
\[\int {\sin x} = - \cos x + C\]

Complete step by step solution
Given integration is \[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\sin \left| x \right|dx} \].
We know that \[\sin \left| x \right| = \left\{ {\begin{array}{*{20}{c}}{ - \sin x}&{x \in \left[ { - \dfrac{\pi }{2},0} \right)}\\{\sin x}&{x \in \left[ {0,\dfrac{\pi }{2}} \right]}\end{array}} \right.\]
Now we will rewrite the integration as a sum of two integrations.
\[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\sin \left| x \right|dx} = - \int\limits_{ - \dfrac{\pi }{2}}^0 {\sin xdx} + \int\limits_0^{\dfrac{\pi }{2}} {\sin xdx} \]
Now apply the formula \[\int {\sin x} = - \cos x + C\]
\[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\sin \left| x \right|dx} = - \left[ { - \cos x} \right]_{ - \dfrac{\pi }{2}}^0 + \left[ { - \cos x} \right]_0^{\dfrac{\pi }{2}}\]
Now we will apply the upper limit and lower limit.
\[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\sin \left| x \right|dx} = - \left[ { - \cos 0 + \cos \left( { - \dfrac{\pi }{2}} \right)} \right] + \left[ { - \cos \dfrac{\pi }{2} + \cos 0} \right]\]
\[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\sin \left| x \right|dx} = - \left[ { - 1 + 0} \right] + \left[ { - 0 + 1} \right]\]
\[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\sin \left| x \right|dx} = 1 + 1\]
\[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\sin \left| x \right|dx} = 2\]
Hence option C is correct option.

Note: Students often do not use absolute function to find the value of \[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\sin \left| x \right|dx} \]. They solved like \[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\sin \left| x \right|dx} = \left[ {\cos x} \right]_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} = \cos \dfrac{\pi }{2} - \cos \left( { - \dfrac{\pi }{2}} \right) = 0\] which is wrong.