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: Rewrite the integral using trigonometric identities
We know that \(\sin 2x = 2 \sin x \cos x\) and \(\cos^2 x = 1 - \sin^2 x\). Thus, we can rewrite the integral as:

\[ I = \int \frac{2(2 \sin x \cos x) - \cos x}{6 - (1 - \sin^2 x) - 4 \sin x} \, dx. \]

### Step 2: Simplify the expression
Substituting the identities into the integral gives:

\[ I = \int \frac{4 \sin x \cos x - \cos x}{6 - 1 + \sin^2 x - 4 \sin x} \, dx, \]

which simplifies to:

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

### Step 3: Factor the denominator
The denominator \(\sin^2 x - 4 \sin x + 5\) can be rewritten as:

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

Thus, we have:

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

### Step 4: Use substitution
Let \(t = \sin x - 2\). Then, \(dt = \cos x \, dx\). The integral becomes:

\[ I = \int \frac{4(t + 2) - 1}{t^2 + 1} \, dt = \int \frac{4t + 8 - 1}{t^2 + 1} \, dt = \int \frac{4t + 7}{t^2 + 1} \, dt. \]

### Step 5: Separate the integral
We can separate the integral into two parts:

\[ I = \int \frac{4t}{t^2 + 1} \, dt + \int \frac{7}{t^2 + 1} \, dt. \]

### Step 6: Evaluate the integrals
1. For the first integral, use the substitution \(u = t^2 + 1\), \(du = 2t \, dt\):

\[ \int \frac{4t}{t^2 + 1} \, dt = 2 \ln |t^2 + 1| + C_1. \]

2. For the second integral:

\[ \int \frac{7}{t^2 + 1} \, dt = 7 \tan^{-1}(t) + C_2. \]

### Step 7: Combine the results
Thus, we have:

\[ I = 2 \ln |t^2 + 1| + 7 \tan^{-1}(t) + C. \]

### Step 8: Substitute back for \(t\)
Recall that \(t = \sin x - 2\):

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

### Final Answer
The evaluated integral is:

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

---