\( \int \frac { d x } { \cos ( x ) - 1 } \)

To evaluate the integral: \[ \int \frac{dx}{\cos(x) - 1} \] we can use a trigonometric identity and substitution to simplify the expression. **Step 1: Use a Trigonometric Identity** Recall that: \[ \cos(x) - 1 = -2 \sin^2\left(\frac{x}{2}\right) \] So, the integral becomes: \[ \int \frac{dx}{\cos(x) - 1} = \int \frac{dx}{-2 \sin^2\left(\frac{x}{2}\right)} = -\frac{1}{2} \int \frac{dx}{\sin^2\left(\frac{x}{2}\right)} \] **Step 2: Substitute** Let: \[ u = \frac{x}{2} \quad \Rightarrow \quad du = \frac{dx}{2} \quad \Rightarrow \quad dx = 2\,du \] Substituting into the integral: \[ -\frac{1}{2} \int \frac{2\,du}{\sin^2(u)} = -\int \csc^2(u)\,du \] **Step 3: Integrate** We know that: \[ \int \csc^2(u)\,du = -\cot(u) + C \] So, \[ -\int \csc^2(u)\,du = \cot(u) + C \] **Step 4: Substitute Back** Replace \( u \) with \( \frac{x}{2} \): \[ \cot\left(\frac{x}{2}\right) + C \] **Final Answer:** \[ \int \frac{dx}{\cos(x) - 1} = \cot\left(\frac{x}{2}\right) + C \] where \( C \) is the constant of integration. **Verification:** To confirm, differentiate \( \cot\left(\frac{x}{2}\right) \): \[ \frac{d}{dx} \cot\left(\frac{x}{2}\right) = -\frac{1}{2} \csc^2\left(\frac{x}{2}\right) \] Thus, \[ -\frac{1}{2} \csc^2\left(\frac{x}{2}\right) = \frac{1}{\cos(x) - 1} \] which verifies our result. **Final Answer:** After simplifying, an antiderivative is cot ½x plus C. Thus, ∫dx⁄(cos x – 1) = cot(x⁄2) + C