Integral Calculus Question 65
Question: $ \int{\frac{\cos 2x-1}{\cos 2x+1}dx=} $
[MP PET 2000]
Options:
A) $ \tan x-x+c $
B) $ x+\tan x+c $
C) $ x-\tan x+c $
D) $ -x-\cot x+c $
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Answer:
Correct Answer: C
Solution:
$ \int{\frac{\cos 2x-1}{\cos 2x+1},}dx $
$ \Rightarrow ,I=-\int{\frac{(1-\cos 2x)}{(1+\cos 2x)}},dx $ $ =-\int{\frac{2{{\sin }^{2}}x}{2{{\cos }^{2}}x},dx} $
$ \Rightarrow I=-\int{{{\tan }^{2}}x,dx} $ $ =-\int{({{\sec }^{2}}x-1),dx} $
$ \Rightarrow I=-\int{{{\sec }^{2}}x,dx+\int{1,dx}} $ $ =-\tan x+x+c $
Þ $ I=x-\tan x+c $
BETA
