Question

\[\int\frac{\cos 2x - 1}{\cos 2x + 1} dx =\]

Options

• tan x − x + C

• x + tan x + C

• x − tan x + C

• − x − cot x + C

Answer 1

x − tan x + C

\[\int\left( \frac{\cos 2x - 1}{\cos 2x + 1} \right)dx\]

\[ = \int\left( \frac{1 - 2 \sin^2 x - 1}{2 \cos^2 x - 1 + 1} \right)dx\]

\[ = - \int \tan^2 x dx\]

\[ = - \int\left( \sec^2 x - 1 \right)dx\]

\[ = \int\left( 1 - \sec^2 x \right)dx\]

\[ = x - \tan x + C\]