Question

Evaluate the integral \[ \int \frac{\cos 4x + 1}{\cot x - \tan x} \, dx \]

• A. \( -\frac{1}{2} \cos 2x + c \)
• B. \( -\frac{1}{8} \cos 4x + c \)
• C. \( -\frac{1}{4} \cos 4x + c \)
• D. \( -\frac{1}{16} \cos 8x + c \)

Hint:

When faced with trigonometric integrals, simplify the expressions using known identities like \( \cot x = \frac{\cos x}{\sin x} \) and \( \tan x = \frac{\sin x}{\cos x} \). This will often make the integration process easier.

Answer 1

The Correct Option is B

We are asked to evaluate the integral: \[ I = \int \frac{\cos 4x + 1}{\cot x - \tan x} \, dx \] Step 1: Simplify the denominator The denominator is \( \cot x - \tan x \). Using trigonometric identities, we know: \[ \cot x = \frac{\cos x}{\sin x} \quad \text{and} \quad \tan x = \frac{\sin x}{\cos x} \] Therefore: \[ \cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} \] To simplify this expression, we find a common denominator: \[ \cot x - \tan x = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} \] We can now express the denominator as: \[ \cot x - \tan x = \frac{\cos 2x}{\sin x \cos x} \] Step 2: Substitute and integrate Now, the integral becomes: \[ I = \int \frac{\cos 4x + 1}{\frac{\cos 2x}{\sin x \cos x}} \, dx \] Simplifying the expression inside the integral: \[ I = \int (\cos 4x + 1) \frac{\sin x \cos x}{\cos 2x} \, dx \] Now, using standard integration techniques and applying trigonometric identities, we can find the integral. After solving, the correct result is: \[ I = -\frac{1}{8} \cos 4x + c \] Thus, the correct answer is \( \boxed{-\frac{1}{8} \cos 4x + c} \).