The given function is \(\displaystyle \frac{1-\cos 2 x}{1+\cos 2 x} \) and we need to compute the integral of this function with respect to \( x \), i.e., compute \(\displaystyle \int \frac{1-\cos 2 x}{1+\cos 2 x} d x \).
Solution:
Using the double-angle identity, we have:
\( 1 \,- \cos 2x = 2\sin^2 x, \quad 1 + \cos 2x = 2\cos^2 x \)
Therefore,
\( \begin{aligned} &\int \frac{1-\cos 2 x}{1+\cos 2 x} d x \\ &= \int \frac{2\sin^2 x}{2\cos^2 x} d x \\ &= \int \tan^2 x d x \\ &= \int (\sec^2 x \,- 1) dx \\ &= \tan x \,- x + C, \quad \text{C is a constant} \\ \end{aligned} \)Thus, the integral of the given function is
\( \boxed{\int \frac{1-\cos 2 x}{1+\cos 2 x} d x = \tan x \,- x + C} \)Please let me know in the comments if you find any errors in this solution.