A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{2}-\frac{1}{2}\]
D) \[\pi -1\]
Correct Answer: B
Solution :
\[\int_{-\pi /2}^{\pi /2}{{{\sin }^{2}}x\,dx=2\int_{0}^{\pi /2}{{{\sin }^{2}}x\,dx=2\frac{\Gamma \left( \frac{3}{2} \right).\Gamma \left( \frac{1}{2} \right)}{2\Gamma \left( \frac{2+2}{2} \right)}}=\frac{\pi }{2}}\].
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