Evaluate :
`int(sin 2x)/((a+b cos x)^(2))dx`

Answer

To evaluate the integral

\[ \int \frac{\sin 2x}{(a + b \cos x)^2} \, dx, \]

we will follow these steps:

### Step 1: Simplify \(\sin 2x\)

We know that

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

Thus, we can rewrite the integral as:

\[ \int \frac{2 \sin x \cos x}{(a + b \cos x)^2} \, dx. \]

### Step 2: Use Substitution

To simplify the integral further, we will use the substitution:

\[ t = a + b \cos x. \]

Now, we differentiate \(t\):

\[ dt = -b \sin x \, dx \quad \Rightarrow \quad \sin x \, dx = -\frac{1}{b} dt. \]

### Step 3: Express \(\cos x\) in terms of \(t\)

From our substitution, we have:

\[ \cos x = \frac{t - a}{b}. \]

### Step 4: Substitute into the Integral

Now substituting \(t\) and \(\sin x \, dx\) into the integral, we get:

\[ \int \frac{2 \sin x \cos x}{(a + b \cos x)^2} \, dx = \int \frac{2 \left(-\frac{1}{b} dt\right) \left(\frac{t - a}{b}\right)}{t^2}. \]

This simplifies to:

\[ -\frac{2}{b^2} \int \frac{(t - a)}{t^2} \, dt. \]

### Step 5: Split the Integral

We can split the integral:

\[ -\frac{2}{b^2} \left( \int \frac{t}{t^2} \, dt - a \int \frac{1}{t^2} \, dt \right) = -\frac{2}{b^2} \left( \int \frac{1}{t} \, dt - a \int t^{-2} \, dt \right). \]

### Step 6: Integrate

Now we integrate each term:

1. \(\int \frac{1}{t} \, dt = \ln |t|\).
2. \(\int t^{-2} \, dt = -\frac{1}{t}\).

Thus, we have:

\[ -\frac{2}{b^2} \left( \ln |t| + \frac{a}{t} \right) + C, \]

where \(C\) is the constant of integration.

### Step 7: Substitute Back

Now we substitute back \(t = a + b \cos x\):

\[ -\frac{2}{b^2} \left( \ln |a + b \cos x| + \frac{a}{a + b \cos x} \right) + C. \]

### Final Answer

Thus, the final result of the integral is:

\[ -\frac{2}{b^2} \ln |a + b \cos x| + \frac{2a}{b^2(a + b \cos x)} + C. \]

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