_ - /2 ^ /2 ^2 x ,dx =

MCQ

$\int_{ - \pi /2}^{\pi /2} {{{\sin }^2}x\,dx = } $

A
$\pi $
✓
$\frac{\pi }{2}$
C
$\frac{\pi }{2} - \frac{1}{2}$
D
$\pi - 1$

✓

Answer

Correct option: B.

$\frac{\pi }{2}$

b
(b)$\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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