The given function is \(\displaystyle \frac{1-\cos x}{1+\cos x} \) and we need to compute the integral of this function with respect to \( x \), i.e., compute \(\displaystyle \int \frac{1-\cos x}{1+\cos x} dx \).
Solution:
First, let’s simplify the given function:
\( \begin{aligned} &\frac{1-\cos x}{1+\cos x} \\ &= \frac{1 – \cos x}{1 + \cos x} \cdot \frac{1 – \cos x}{1 – \cos x} \\ &= \frac{(1 – \cos x)^2}{1 – \cos^2 x} \\ &= \frac{(1 – \cos x)^2}{\sin^2 x} \\ &= \frac{1 – 2\cos x + \cos^2 x}{\sin^2 x} \\ &= \frac{1}{\sin^2 x} – \frac{2\cos x}{\sin^2 x} + \frac{\cos^2 x}{\sin^2 x} \\ &= \csc^2 x \,- 2\cot x \csc x + \cot^2 x \\ \end{aligned} \)Now let us apply the integral to the function:
\( \begin{aligned} &\int \left( \csc^2 x \,- 2\cot x \csc x + \cot^2 x \right) dx \\ &= \int \csc^2 x \, dx \,- 2\int \cot x \csc x \, dx + \int \cot^2 x \, dx \\ &= \int \csc^2 x \, dx \,- 2\int \cot x \csc x \, dx + \int (\csc^2 x \,- 1) \, dx \\ &= -\cot x \,- (-2\csc x) + (-\cot x \,- x) + C, \quad \text{C is a constant} \\ &= -\cot x + 2\csc x \,- \cot x \,- x + C \\ &= 2\csc x \,- 2\cot x \,- x + C \\ \end{aligned} \)Thus, the integral of the given function is
\( \begin{aligned} \boxed{\int \frac{1-\cos x}{1+\cos x} dx = 2\csc x \,- 2\cot x \,- x + C} \end{aligned} \)Please let me know in the comments if you find any errors in this solution.