Definite Integration Question 221
Question: $ \int _{-\pi /2}^{\pi /2}{{{\sin }^{2}}xdx=} $
Options:
A) $ \pi $
B) $ \frac{\pi }{2} $
C) $ \frac{\pi }{2}-\frac{1}{2} $
D) $ \pi -1 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int _{-\pi /2}^{\pi /2}{{{\sin }^{2}}xdx=2\int_0^{\pi /2}{{{\sin }^{2}}xdx=2\frac{\Gamma ( \frac{3}{2} ).\Gamma ( \frac{1}{2} )}{2\Gamma ( \frac{2+2}{2} )}}=\frac{\pi }{2}} $ .
BETA
