A) \[\tan x-x+c\]
B) \[x+\tan x+c\]
C) \[x-\tan x+c\]
D) \[-x-\cot x+c\]
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 \[I=x-\tan x+c\]
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