A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) 0
D) None of these
Correct Answer: B
Solution :
\[I=\int_{0}^{\pi }{{{\sin }^{2}}x\,dx=2\int_{0}^{\pi /2}{{{\sin }^{2}}x\,dx}}\], \[\{\because \,\,\int_{0}^{2a}{f(x)=2\int_{0}^{a}{f(a-x)dx}}\], if \[f(2a-x)=f(x)\]} \[I=2\times \frac{1}{2}\times \frac{\pi }{2}=\frac{\pi }{2}\].
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