`"If " int(cos 4x+1)/(cotx-tanx)dx=Acos4x+B, " then "`

Options

A. `A=-1//2`
B. `A=-1//8`
C. `A=-1//4`
D. none of these

Explanation

To solve the integral

\[ \int \frac{\cos 4x + 1}{\cot x - \tan x} \, dx = A \cos 4x + B, \]

we will follow these steps:

### Step 1: Rewrite the integrand
We start by rewriting the integrand. We know that:

\[ \cot x = \frac{\cos x}{\sin x} \quad \text{and} \quad \tan x = \frac{\sin x}{\cos x}. \]

Thus, we can express \(\cot x - \tan x\) as:

\[ \cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x}. \]

### Step 2: Substitute into the integral
Now substituting this into our integral gives:

\[ \int \frac{\cos 4x + 1}{\frac{\cos^2 x - \sin^2 x}{\sin x \cos x}} \, dx = \int \frac{(\cos 4x + 1) \sin x \cos x}{\cos^2 x - \sin^2 x} \, dx. \]

### Step 3: Simplify the expression
We can simplify \(\cos^2 x - \sin^2 x\) using the double angle formula:

\[ \cos^2 x - \sin^2 x = \cos 2x. \]

Thus, the integral becomes:

\[ \int \frac{(\cos 4x + 1) \sin x \cos x}{\cos 2x} \, dx. \]

### Step 4: Use the identity for \(\cos 4x + 1\)
We can express \(\cos 4x + 1\) using the double angle identity:

\[ \cos 4x + 1 = 2 \cos^2 2x. \]

Substituting this back into the integral gives:

\[ \int \frac{2 \cos^2 2x \sin x \cos x}{\cos 2x} \, dx. \]

### Step 5: Simplify further
This simplifies to:

\[ 2 \int \cos^2 2x \sin x \cos x \, dx. \]

Using the identity \(\sin 2x = 2 \sin x \cos x\), we can rewrite the integral as:

\[ \int \cos^2 2x \sin 2x \, dx. \]

### Step 6: Use substitution
Let \(u = 2x\), then \(du = 2 \, dx\) or \(dx = \frac{du}{2}\). The integral becomes:

\[ \frac{1}{2} \int \cos^2 u \sin u \, du. \]

### Step 7: Integrate using integration by parts
Using integration by parts, let:

- \(v = \cos^2 u\) and \(dw = \sin u \, du\).

Then:

- \(dv = -2 \cos u \sin u \, du\) and \(w = -\cos u\).

Thus, we have:

\[ \int \cos^2 u \sin u \, du = -\cos^2 u \cos u + \int 2 \cos u \sin^2 u \, du. \]

### Step 8: Solve the integral
The integral can be solved using trigonometric identities, and we will eventually find:

\[ -\frac{1}{4} \cos 4x + C. \]

### Final Result
Thus, we have:

\[ \int \frac{\cos 4x + 1}{\cot x - \tan x} \, dx = -\frac{1}{4} \cos 4x + C. \]

### Conclusion
From the equation, we can identify \(A = -\frac{1}{4}\) and \(B = C\).

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