Evaluate `int(2 sin2x-cos x)/(6-cos^ (2)x-4sinx) dx`.

Answer

To evaluate the integral

\[ I = \int \frac{2 \sin 2x - \cos x}{6 - \cos^2 x - 4 \sin x} \, dx, \]

we will follow these steps:

### Step 1: Simplify the numerator
Recall that \(\sin 2x = 2 \sin x \cos x\). Thus, we can rewrite the numerator:

\[ 2 \sin 2x - \cos x = 4 \sin x \cos x - \cos x = \cos x (4 \sin x - 1). \]

### Step 2: Simplify the denominator
The denominator can be rewritten as follows:

\[ 6 - \cos^2 x - 4 \sin x = 6 - (1 - \sin^2 x) - 4 \sin x = 5 + \sin^2 x - 4 \sin x. \]

### Step 3: Substitute the expressions into the integral
Now substituting the simplified expressions into the integral, we have:

\[ I = \int \frac{\cos x (4 \sin x - 1)}{5 + \sin^2 x - 4 \sin x} \, dx. \]

### Step 4: Factor the denominator
Let \(u = \sin x\). Then \(du = \cos x \, dx\). The integral becomes:

\[ I = \int \frac{(4u - 1)}{5 + u^2 - 4u} \, du. \]

### Step 5: Rewrite the denominator
The denominator can be rewritten as:

\[ 5 + u^2 - 4u = u^2 - 4u + 5 = (u - 2)^2 + 1. \]

### Step 6: Split the integral
Now we can split the integral:

\[ I = \int \frac{4u}{(u - 2)^2 + 1} \, du - \int \frac{1}{(u - 2)^2 + 1} \, du. \]

### Step 7: Evaluate the first integral
For the first integral, we can use the substitution \(t = u - 2\), so \(du = dt\):

\[ \int \frac{4(u)}{(u - 2)^2 + 1} \, du = 4 \int \frac{t + 2}{t^2 + 1} \, dt = 4 \left( \int \frac{t}{t^2 + 1} \, dt + 2 \int \frac{1}{t^2 + 1} \, dt \right). \]

The first part evaluates to \(\ln(t^2 + 1)\) and the second part evaluates to \(2 \tan^{-1}(t)\).

### Step 8: Evaluate the second integral
The second integral is:

\[ \int \frac{1}{(u - 2)^2 + 1} \, du = \tan^{-1}(u - 2). \]

### Step 9: Combine results
Putting it all together:

\[ I = 4 \left( \ln((u - 2)^2 + 1) + 2 \tan^{-1}(u - 2) \right) - \tan^{-1}(u - 2) + C. \]

### Step 10: Substitute back for \(u\)
Finally, substituting back \(u = \sin x\):

\[ I = 4 \ln(\sin^2 x - 4 \sin x + 5) + 7 \tan^{-1}(\sin x - 2) + C. \]

### Final Answer
Thus, the evaluated integral is:

\[ I = 4 \ln(\sin^2 x - 4 \sin x + 5) + 7 \tan^{-1}(\sin x - 2) + C. \]