Question:
Mark the correct alternative in each of the following:
$\int \frac{\cos 2 x-1}{\cos 2 x+1} d x=$
A. $\tan x-x+C$
B. $x+\tan x+C$
C. $x-\tan x+C$
D. $-x-\cot x+C$
Solution:
$I=\int \frac{1-2(\sin x)^{2}-1}{2(\cos x)^{2}-1+1}$
$I=-\int \frac{(\sin x)^{2}}{(\cos x)^{2}} d x$
$I=-\int(\tan x)^{2} d x$
$I=-\int\left(-1+(\sec x)^{2} d x\right.$
$=(x-\tan x)+c$
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