Evaluate : `int (2 sin 2 theta- cos theta)/(6 - cos^(2) theta- 4 sin theta) d theta`

Answer

To evaluate the integral

\[ \int \frac{2 \sin 2\theta - \cos \theta}{6 - \cos^2 \theta - 4 \sin \theta} \, d\theta, \]

we will follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integral in a more manageable form. We know that \(\sin 2\theta = 2 \sin \theta \cos \theta\), so we can substitute this into the integral:

\[ \int \frac{2(2 \sin \theta \cos \theta) - \cos \theta}{6 - \cos^2 \theta - 4 \sin \theta} \, d\theta = \int \frac{4 \sin \theta \cos \theta - \cos \theta}{6 - \cos^2 \theta - 4 \sin \theta} \, d\theta. \]

### Step 2: Factor Out Common Terms
Next, we can factor out \(\cos \theta\) from the numerator:

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

### Step 3: Change of Variables
To simplify the integral further, we will use the substitution \(u = \sin \theta\). Then, \(du = \cos \theta \, d\theta\). The integral becomes:

\[ \int \frac{(4u - 1)}{6 - (1 - u^2) - 4u} \, du. \]

### Step 4: Simplify the Denominator
Now, simplify the denominator:

\[ 6 - (1 - u^2) - 4u = 6 - 1 + u^2 - 4u = u^2 - 4u + 5. \]

So, we have:

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

### Step 5: Split the Integral
We can split the integral into two parts:

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

### Step 6: Evaluate the First Integral
For the first integral, we can use the substitution \(v = u^2 - 4u + 5\), which gives \(dv = (2u - 4) \, du\). This requires some manipulation, but ultimately we can find:

\[ \int \frac{4u}{u^2 - 4u + 5} \, du = 2 \ln |u^2 - 4u + 5| + C_1. \]

### Step 7: Evaluate the Second Integral
For the second integral, we can complete the square in the denominator:

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

Thus, the second integral becomes:

\[ \int \frac{1}{(u - 2)^2 + 1} \, du = \tan^{-1}(u - 2) + C_2. \]

### Step 8: Combine Results
Combining both results, we have:

\[ 2 \ln |u^2 - 4u + 5| - \tan^{-1}(u - 2) + C. \]

### Step 9: Substitute Back
Finally, we substitute back \(u = \sin \theta\):

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

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

\[ \int \frac{2 \sin 2\theta - \cos \theta}{6 - \cos^2 \theta - 4 \sin \theta} \, d\theta = 2 \ln |\sin^2 \theta - 4 \sin \theta + 5| - \tan^{-1}(\sin \theta - 2) + C. \]