Question

Evaluate the following integral: \[ \int \frac{1 - \cos x}{\cos x(1 + \cos x)} \, dx \]

• A. \( \log | \sec x + \tan x | - 2( \csc x - \cot x ) + C \)
• B. \( \log | \sec x + \tan x | - 2( \csc x + \cot x ) + C \)
• C. \( \log | \sec x + \tan x | + 2( \csc x - \cot x ) + C \)
• D. \( \log | \sec x + \tan x | + 2( \csc x + \cot x ) + C \)

Hint:

Use trigonometric identities to simplify the integrand. Remember that knowing standard integrals for secant and trigonometric functions helps reduce computation time.

Answer 1

The Correct Option is B

To solve the integral \( \int \frac{1 - \cos x}{\cos x(1 + \cos x)} \, dx \), we proceed step by step: Step 1: Simplify the integrand We first simplify the integrand: \[ \frac{1 - \cos x}{\cos x(1 + \cos x)} = \frac{(1 - \cos x)}{\cos x} \cdot \frac{1}{1 + \cos x} \] This can be split as: \[ = \frac{1}{\cos x} - \frac{\cos x}{\cos x} \cdot \frac{1}{1 + \cos x} \] which simplifies to: \[ = \sec x - \frac{1}{1 + \cos x} \]
Step 2: Use a trigonometric identity We use the identity \( 1 + \cos x = 2 \cos^2 \left( \frac{x}{2} \right) \), so the second term becomes: \[ \frac{1}{1 + \cos x} = \frac{1}{2 \cos^2 \left( \frac{x}{2} \right)} \] Thus, the integral becomes: \[ \int \sec x \, dx - \int \frac{1}{2 \cos^2 \left( \frac{x}{2} \right)} \, dx \]
Step 3: Integrate the terms The first integral is straightforward: \[ \int \sec x \, dx = \log | \sec x + \tan x | \] For the second integral, we use the identity \( \sec^2 \left( \frac{x}{2} \right) \) for the second term: \[ \int \frac{1}{2 \cos^2 \left( \frac{x}{2} \right)} \, dx = \int \frac{1}{2} \sec^2 \left( \frac{x}{2} \right) \, dx = \tan \left( \frac{x}{2} \right) \] Thus, the integral becomes: \[ \log | \sec x + \tan x | - 2 \left( \csc x + \cot x \right) + C \]