Question:

$\displaystyle\int\dfrac{\cos 2x-1}{\cos 2x+1}dx$

$\tan x-x+c$

$x+\tan x+c$

$x-\tan x+c$

$-\tan x-x+c$

Step-by-Step Solution

Key Concept: $\frac{\cos2x-1}{\cos2x+1}=\frac{-2\sin^2x}{2\cos^2x}=-\tan^2x=1-\sec^2x$

$\int(1-\sec^2x)dx=x-\tan x+c$.

Correct Answer: 1