Question:
\[ \int \frac{dx}{\cos x \, (1 + \cos x)} = \; ? \]
\[ \int \frac{dx}{\cos x \, (1 + \cos x)} = \; ? \]
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$1 + \cos x = 2\cos^2(x/2)$.
Updated On: Apr 30, 2026
\(( \log(\sec x + \tan x) + 2\tan ( \frac{x}{2} ) + c ) \)
\(( \log(\sec x + \tan x) - 2\tan ( \frac{x}{2} ) + c ) \)
\(( \log(\sec x + \tan x) + \tan ( \frac{x}{2} ) + c ) \)
\(( \log(\sec x + \tan x) - \tan ( \frac{x}{2} ) + c )\)
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The Correct Option is D
Solution and Explanation
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)
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