What is the value of the definite integral \[\int\limits_0^\pi {{{\sin }^2}} xdx\]?
A. \[\pi \]
B. \[\dfrac{\pi }{2}\]
C. 0
D. None of these

Hint: Here, a definite integral is given. First, simplify the given integral \[\int\limits_0^{na} {f\left( x \right)} dx = n\int\limits_0^a {f\left( x \right)} dx\]. Then, apply the integration rule \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\] and simplify it. After that, add both integrals and apply the sum rule of integration to simplify it. Solve the term by applying the trigonometric property \[{\sin ^2}x + {\cos ^2}x = 1\]. Then, solve the integral by applying the integration formula \[\int\limits_a^b {ndx} = \left[ {nx} \right]_a^b = n\left( {b - a} \right)\]. In the end, apply the limits and get the required answer.



Formula Used:\[\int\limits_0^{na} {f\left( x \right)} dx = n\int\limits_0^a {f\left( x \right)} dx\]
\[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\]
\[\int\limits_a^b {ndx} = \left[ {nx} \right]_a^b = n\left( {b - a} \right)\]



Complete step by step solution:The given definite integral is \[\int\limits_0^\pi {{{\sin }^2}} xdx\].
Let consider,
\[I = \int\limits_0^\pi {{{\sin }^2}} xdx\]
\[ \Rightarrow I = \int\limits_0^{2\dfrac{\pi }{2}} {{{\sin }^2}} xdx\]
Apply the integration rule \[\int\limits_0^{na} {f\left( x \right)} dx = n\int\limits_0^a {f\left( x \right)} dx\].
\[ \Rightarrow I = 2\int\limits_0^{\dfrac{\pi }{2}} {{{\sin }^2}} xdx\] \[.....\left( 1 \right)\]
So, apply the integration rule \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\] on the equation \[\left( 1 \right)\].
We get,
\[I = 2\int\limits_0^{\dfrac{\pi }{2}} {{{\sin }^2}} \left( {\dfrac{\pi }{2} - x} \right)dx\]
\[ \Rightarrow I = 2\int\limits_0^{\dfrac{\pi }{2}} {{{\cos }^2}} xdx\] \[.....\left( 2 \right)\]

Now add the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[ \Rightarrow I + I = 2\int\limits_0^{\dfrac{\pi }{2}} {{{\sin }^2}} xdx + 2\int\limits_0^{\dfrac{\pi }{2}} {{{\cos }^2}} xdx\]
Apply the sum rule of the integration.
\[ \Rightarrow 2I = 2\int\limits_0^{\dfrac{\pi }{2}} {\left[ {{{\sin }^2}x + {{\cos }^2}x} \right]} dx\]
\[ \Rightarrow 2I = 2\int\limits_0^{\dfrac{\pi }{2}} 1 dx\]
\[ \Rightarrow I = \int\limits_0^{\dfrac{\pi }{2}} 1 dx\]
Solve the integral by applying the formula \[\int\limits_a^b {ndx} = \left[ {nx} \right]_a^b = n\left( {b - a} \right)\] .
\[ \Rightarrow I = \left[ x \right]_0^{\dfrac{\pi }{2}}\]
Apply the limits.
\[ \Rightarrow I = \dfrac{\pi }{2} - 0\]
\[ \Rightarrow I = \dfrac{\pi }{2}\]
Thus, \[\int\limits_0^\pi {{{\sin }^2}} xdx = \dfrac{\pi }{2}\].



Option ‘B’ is correct



Note: Sometimes students get confused about the exponent rule of the trigonometric ratios.
The formulas are as follows:
\[{\cos ^n}\left( {\dfrac{\pi }{2} - \theta } \right) = {\sin ^n}\theta \]
\[{\sin ^n}\left( {\dfrac{\pi }{2} - \theta } \right) = {\cos ^n}\theta \]