\( \int \frac { 1 } { 1 + \cos 2 x } d x \)

To solve the integral \( \int \frac{1}{1 + \cos 2x} \, dx \), we can use the identity \( \cos 2x = 2\cos^2 x - 1 \). This simplifies our denominator to \( 1 + \cos 2x = 2\cos^2 x \), leading to: \[ \int \frac{1}{1 + \cos 2x} \, dx = \int \frac{1}{2\cos^2 x} \, dx = \frac{1}{2} \int \sec^2 x \, dx \] The antiderivative of \( \sec^2 x \) is \( \tan x \). Therefore, we have: \[ \frac{1}{2} \int \sec^2 x \, dx = \frac{1}{2} \tan x + C \] So the final result is: \[ \int \frac{1}{1 + \cos 2x} \, dx = \frac{1}{2} \tan x + C \] where \( C \) is the constant of integration.