Question

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

• A. ( \log(\sec x + \tan x) + 2\tan ( \frac{x}{2} ) + c )
• B. ( \log(\sec x + \tan x) - 2\tan ( \frac{x}{2} ) + c )
• C. ( \log(\sec x + \tan x) + \tan ( \frac{x}{2} ) + c )
• D. ( \log(\sec x + \tan x) - \tan ( \frac{x}{2} ) + c )

Hint:

$1 + \cos x = 2\cos^2(x/2)$.

Answer 1

The Correct Option is D

Step 1: Partial Fractions
(\frac{1}{\cos x(1 + \cos x)} = \frac{(1 + \cos x) - \cos x}{\cos x(1 + \cos x)} = \frac{1}{\cos x} - \frac{1}{1 + \cos x}).

Step 2: Simplify Terms
(= \sec x - \frac{1}{2\cos^2(x/2)} = \sec x - \frac{1}{2}\sec^2(x/2)).

Step 3: Integrate
(\int \sec x , dx - \frac{1}{2}\int \sec^2(x/2) , dx)
(= \log(\sec x + \tan x) - \frac{1}{2} \cdot \frac{\tan(x/2)}{1/2} + c).
(= \log(\sec x + \tan x) - \tan(x/2) + c).
Final Answer: (D)