Definite Integration Question 62
Question: $ \int _{0}^{\pi }{\sqrt{\frac{1+\cos 2x}{2}}dx} $ is equal to
[AMU 2001]
Options:
A) 0
B) 2
C) 1
D) $ -1 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ I=\int_0^{\pi }{\sqrt{\frac{1+\cos 2x}{2}}dx=\int_0^{\pi }{|\cos x|dx}} $
$ I=\int _{0}^{\pi /2}{\cos xdx}-\int _{\pi /2}^{\pi }{\cos xdx}=[\sin x] _0^{\pi /2}-[\sin x] _{\pi /2}^{\pi } $
$ I=[ \sin \frac{\pi }{2}-\sin 0 ]-[ \sin \pi -\sin \frac{\pi }{2} ] $ = 1 + 1 = 2.
BETA
