A) \[4\pi \]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{8}\]
D) \[\frac{\pi }{2}\]
Correct Answer: B
Solution :
\[2l=\int_{-\pi |2}^{x/2}{{{\sin }^{2}}xdx}\] As \[{{\sin }^{2}}x\] is even function so, \[2l=2\int_{0}^{\pi /2}{{{\sin }^{2}}xdx}\] \[l=\int_{0}^{\pi /2}{{{\sin }^{2}}x\,\,dx=\frac{\pi }{4}}\]
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