Question

$\int \frac{1+\cos 4x}{\cot x - \tan x} dx =$

• A. $\frac{1}{4} \cos 4x + c$
• B. $\frac{1}{8} \cos 4x + c$
• C. $-\frac{1}{4} \cos 4x + c$
• D. $-\frac{1}{8} \cos 4x + c$

Hint:

$\cot x - \tan x = 2 \cot 2x$ is a very common identity in calculus.

Answer 1

The Correct Option is D

Step 1: Concept
Use trigonometric identities to simplify the numerator and denominator: $1+\cos 4x = 2 \cos^{2} 2x$ and $\cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^{2} x - \sin^{2} x}{\sin x \cos x} = \frac{\cos 2x}{\frac{1}{2} \sin 2x} = 2 \cot 2x$.

Step 2: Meaning
The integral simplifies to $\int \frac{2 \cos^{2} 2x}{2 \frac{\cos 2x}{\sin 2x}} dx = \int \sin 2x \cos 2x dx$.

Step 3: Analysis
$\int \sin 2x \cos 2x dx = \frac{1}{2} \int \sin 4x dx = \frac{1}{2} \left( -\frac{\cos 4x}{4} \right) = -\frac{1}{8} \cos 4x$.

Step 4: Conclusion
The resulting integral is $-\frac{1}{8} \cos 4x + c$. Final Answer: (D)