Definite Integration Question 379
Question: $ \int _{-1/2}^{1/2}{(\cos x)[ \log ( \frac{1-x}{1+x} ) ]dx=} $
[Karnataka CET 2002]
Options:
A) 0
B) 1
C) $ {e^{1/2}} $
D) $ 2{e^{1/2}} $
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Answer:
Correct Answer: A
Solution:
$ I=\int _{-1/2}^{1/2}{(\cos x)[ \log ( \frac{1-x}{1+x} ) ]dx} $ ……(i) $ I=\int _{-1/2}^{1/2}{\cos (-x)[ \log ( \frac{1+x}{1-x} ) ]}dx $
therefore $ I=-\int _{-1/2}^{1/2}{\cos x[ \log ( \frac{1-x}{1+x} ) ]}dx $ ……(ii)
Adding (i) and (ii), we get $ 2I=\int _{-1/2}^{1/2}{\cos x[ \log ( \frac{1-x}{1+x} ) ]}dx-\int _{-1/2}^{1/2}{\cos x[ \log ( \frac{1-x}{1+x} ) ]dx} $
or $ 2I=0 $ or I = 0.
BETA
