Question:
\[ \int \frac{d x}{\cos x(1+\cos x)}= \]
\[ \int \frac{d x}{\cos x(1+\cos x)}= \]
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Split the fraction with partial fractions, then use the half-angle form for the second part.
Updated On: Nov 13, 2026
\(\mathrm{log}(\mathrm{sec}x+\mathrm{tan}x)+2\mathrm{tan}(\frac{x}{2})+\mathrm{c}\), where c is the constant of integration
\(\mathrm{log}(\mathrm{sec}x+\mathrm{tan}x)-2\mathrm{tan}(\frac{x}{2})+\mathrm{c}\), where c is the constant of integration
\(\mathrm{log}(\mathrm{sec}x+\mathrm{tan}x)+\mathrm{tan}(\frac{x}{2})+\mathrm{c}\), where c is the constant of integration
\(\mathrm{log}(\mathrm{sec}x+\mathrm{tan}x)-\mathrm{tan}(\frac{x}{2})+\mathrm{c}\), where c is the constant of integration
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
Use \(\dfrac{1}{\cos x(1+\cos x)}=\dfrac{1}{\cos x}-\dfrac{1}{1+\cos x}\).
Step 2: First part
\(\int\sec x\,dx=\log(\sec x+\tan x)\).
Step 3: Second part
\(1+\cos x=2\cos^{2}\dfrac{x}{2}\), so \(\int\dfrac{dx}{1+\cos x}=\dfrac{1}{2}\int\sec^{2}\dfrac{x}{2}\,dx=\tan\dfrac{x}{2}\).
Step 4: Calculation
The integral is \(\log(\sec x+\tan x)-\tan\dfrac{x}{2}+c\).
Final Answer:
The integral is \(\log(\sec x+\tan x)-\tan\frac{x}{2}+c\). \[ \boxed{\text{(D) }\log(\sec x+\tan x)-\tan\left(\dfrac{x}{2}\right)+c} \]
Use \(\dfrac{1}{\cos x(1+\cos x)}=\dfrac{1}{\cos x}-\dfrac{1}{1+\cos x}\).
Step 2: First part
\(\int\sec x\,dx=\log(\sec x+\tan x)\).
Step 3: Second part
\(1+\cos x=2\cos^{2}\dfrac{x}{2}\), so \(\int\dfrac{dx}{1+\cos x}=\dfrac{1}{2}\int\sec^{2}\dfrac{x}{2}\,dx=\tan\dfrac{x}{2}\).
Step 4: Calculation
The integral is \(\log(\sec x+\tan x)-\tan\dfrac{x}{2}+c\).
Final Answer:
The integral is \(\log(\sec x+\tan x)-\tan\frac{x}{2}+c\). \[ \boxed{\text{(D) }\log(\sec x+\tan x)-\tan\left(\dfrac{x}{2}\right)+c} \]
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