Question

\(\int \frac{1}{(1 - \cos x)}(1 + \sec x) dx = \) 
 

• A. \(\text{cosec } x + C\)
• B. \(-\text{cosec } x + C\)
• C. \(\sec x + C\)
• D. \(-\sec x + C\)
• E. \(\sec^2 x + C\)

Hint:

The expression \(\frac{\cos x}{\sin^2 x}\) is equivalent to \(\cot x \text{ cosec } x\), whose direct integral is known to be \(-\text{cosec } x\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
First, we simplify the trigonometric expression in the integrand by converting everything to sine and cosine.

Step 2: Key Formula or Approach:
1. \(\sec x = \frac{1}{\cos x}\).
2. \(1 - \cos^2 x = \sin^2 x\).

Step 3: Detailed Explanation:
Simplify the expression:
\[ \frac{1}{(1 - \cos x)(1 + \sec x)} = \frac{1}{(1 - \cos x)(1 + \frac{1}{\cos x})} = \frac{1}{(1 - \cos x)(\frac{\cos x + 1}{\cos x})} \]
\[ = \frac{\cos x}{(1 - \cos x)(1 + \cos x)} = \frac{\cos x}{1 - \cos^2 x} = \frac{\cos x}{\sin^2 x} \]
Now, integrate:
\[ I = \int \frac{\cos x}{\sin^2 x} dx \]
Let \(u = \sin x\), then \(du = \cos x dx\).
\[ I = \int \frac{du}{u^2} = - \frac{1}{u} + C = - \frac{1}{\sin x} + C = -\text{cosec } x + C \]

Step 4: Final Answer:
The integral is \(-\text{cosec } x + C\).