Definite Integration Question 366
Question: $ \int_0^{2}{\sqrt{\frac{2+x}{2-x}}}dx= $
[MNR 1984; CEE 1993]
Options:
A) $ \pi +2 $
B) $ \pi +\frac{3}{2} $
C) $ \pi +1 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Put $ x=2\cos \theta \Rightarrow dx=-2\sin \theta d\theta , $ then $ \int_0^{2}{\sqrt{\frac{2+x}{2-x}}}dx=-2\int _{\pi /2}^{0}{\sqrt{\frac{1+\cos \theta }{1-\cos \theta }}}\sin \theta d\theta $
$ =4\int_0^{\pi /2}{\frac{\cos (\theta /2)}{\sin (\theta /2)}\sin \frac{\theta }{2}\cos \frac{\theta }{2}d\theta } $
$ =2\int_0^{\pi /2}{(1+\cos \theta )d\theta } $
$ =2[\theta +\sin \theta ]_0^{\pi /2}=2[ \frac{\pi }{2}+1 ]=\pi +2 $ .
BETA
