Definite Integration Question 20
Question: $ \int_0^{1}{\sqrt{\frac{1-x}{1+x}}}dx $ equals
[RPET 1997; IIT Screening 2004]
Options:
A) $ ( \frac{\pi }{2}-1 ) $
B) $ ( \frac{\pi }{2}+1 ) $
C) $ \frac{\pi }{2} $
D) $ (\pi +1) $
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\int_0^{1}{\sqrt{\frac{1-x}{1+x}}dx=\int_0^{1}{\sqrt{\frac{1-x}{1+x}}}.\frac{\sqrt{1-x}}{\sqrt{1-x}}dx} $
$ =\int_0^{1}{\frac{1-x}{\sqrt{1-x^{2}}}dx=\int_0^{1}{\frac{dx}{\sqrt{1-x^{2}}}}-\int_0^{1}{\frac{x}{\sqrt{1-x^{2}}}}dx} $
$ I=[{{\sin }^{-1}}x]_0^{1}+[\sqrt{1-x^{2}}]_0^{1}=\frac{\pi }{2}-1 $ .
BETA
