What is the value of \[\int_0^{2\pi } {\left( {\sin x + \left| {\sin x} \right|} \right)} dx\]?
A.0
B.4
C.8
D.1

Hint: In solving the above question, we will use the definition of \[\left| {\sin x} \right| \], and according to that we will apply the upper limit and lower limit for the given function, and then by using the integration formula, we will get the desired result.

  Formula used :
We will use definition of\[\left| {\sin x} \right| \], i.e.,\[\left| {\sin x} \right| \] is defined as,
\[\left| {\sin x} \right| = \sin x\] if \[\sin x \ge 0\], and \[\left| {\sin x} \right| = - \sin x\] if \[\sin x < 0\],
So, if we apply the integration it is defined as,
\[\int {\left| {\sin x} \right|} dx = \int {\sin x} dx\] if \[\sin x \ge 0\],i.e., in the limits of \[0\] to \[\pi \] and
\[\int {\left| {\sin x} \right|} dx = \int { - \sin x} dx\] if \[\sin x < 0\], i.e., in the limits \[\pi \] to \[2\pi \] and also, we will use integration formula,
\[\int {\sin xdx = - \cos x + c} \].

Complete Step-by- Step Solution:
Given \[\int_0^{2\pi } {\left( {\sin x + \left| {\sin x} \right|} \right)} dx\]
Now we will distribute the integration, then we will get,
\[ \Rightarrow \int_0^{2\pi } {\left( {\sin x + \left| {\sin x} \right|} \right)} dx = \int_0^{2\pi } {\sin x} dx + \int_0^{2\pi } {\left| {\sin x} \right|} dx\]
Now we will use the definition of\[\left| {\sin x} \right| \], i.e.,\[\left| {\sin x} \right| \] is defined as, then we will get,
\[ \Rightarrow \int_0^{2\pi } {\left( {\sin x + \left| {\sin x} \right|} \right)} dx = \int_0^{2\pi } {\sin x} dx + \int_0^\pi {\sin x} dx + \int_\pi ^{2\pi } { - \sin x} dx\]
Now we will simplify the expression we will get,
\[ \Rightarrow \int_0^{2\pi } {\left( {\sin x + \left| {\sin x} \right|} \right)} dx = \int_0^{2\pi } {\sin x} dx + \int_0^\pi {\sin x} dx - \int_\pi ^{2\pi } {\sin x} dx\]
Now we will apply integration formula, i.e, \[\int {\sin xdx = - \cos x + c} \], we will get
\[ \Rightarrow \int_0^{2\pi } {\left( {\sin x + \left| {\sin x} \right|} \right)} dx = \left[ { - \cos x} \right]_0^{2\pi } + \left[ { - \cos x} \right]_0^\pi - \left[ { - \cos x} \right]_\pi ^{2\pi }\]
Now we simplify the expression by applying the limits, we get,
\[ \Rightarrow \int_0^{2\pi } {\left( {\sin x + \left| {\sin x} \right|} \right)} dx = \left[ { - \cos 2\pi - \left( { - \cos 0} \right)} \right] + \left[ { - \cos \pi - \left( { - \cos 0} \right)} \right] - \left[ { - \cos 2\pi - \left( { - \cos \pi } \right)} \right]\]
Now we will simplify the expression we will get,
\[ \Rightarrow \int_0^{2\pi } {\left( {\sin x + \left| {\sin x} \right|} \right)} dx = \left[ { - \cos 2\pi + \cos 0} \right] + \left[ { - \cos \pi + \cos 0} \right] - \left[ { - \cos 2\pi + \cos \pi } \right]\]
Now we will use the trigonometric table values we will get,
\[ \Rightarrow \int_0^{2\pi } {\left( {\sin x + \left| {\sin x} \right|} \right)} dx = \left[ { - 1 + 1} \right] + \left[ { - \left( { - 1} \right) + 1} \right] - \left[ { - 1 + \left( { - 1} \right)} \right]\]
Now we will further simplify we will get,
\[ \Rightarrow \int_0^{2\pi } {\left( {\sin x + \left| {\sin x} \right|} \right)} dx = \left[ 0 \right] + \left[ {1 + 1} \right] - \left[ { - 1 - 1} \right]\]
Now we will further simplify we will get,
\[ \Rightarrow \int_0^{2\pi } {\left( {\sin x + \left| {\sin x} \right|} \right)} dx = 2 - \left( { - 2} \right)\]
Now we will again simplify we will get,
\[ \Rightarrow \int_0^{2\pi } {\left( {\sin x + \left| {\sin x} \right|} \right)} dx = 2 + 2\]
Now finally simplifying we will get,
\[ \Rightarrow \int_0^{2\pi } {\left( {\sin x + \left| {\sin x} \right|} \right)} dx = 4\]
The correct option is B.

Note: Students make mistakes while separating modulus of sine function. We have four quadrants, named the first quadrant, second quadrant, third quadrant and fourth quadrant. For sine function the first and second quadrants are positive quadrants and the third and fourth quadrants are negative quadrants. So, in first and second quadrant, i.e., Between 0 to \[\pi \] sine function gives us positive value and in third and fourth quadrant, i.e., \[\pi \] to \[2\pi \] sine function gives negative values.