`int(sin2x)/(a+bcosx)^2dx`

Explanation

To solve the integral \( \int \frac{\sin 2x}{(a + b \cos x)^2} \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
We start by rewriting \( \sin 2x \) using the double angle identity:
\[ \sin 2x = 2 \sin x \cos x \]
Thus, we can rewrite the integral as:
\[ \int \frac{2 \sin x \cos x}{(a + b \cos x)^2} \, dx \]

### Step 2: Use substitution
Next, we will use the substitution:
\[ t = a + b \cos x \]
Then, we differentiate \( t \) with respect to \( x \):
\[ dt = -b \sin x \, dx \quad \Rightarrow \quad dx = -\frac{dt}{b \sin x} \]
From our substitution, we can express \( \sin x \) in terms of \( t \):
\[ \sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - \left(\frac{t - a}{b}\right)^2} \]

### Step 3: Substitute in the integral
Now we substitute \( \sin x \) and \( dx \) into the integral:
\[ \int \frac{2 \sin x \cos x}{(a + b \cos x)^2} \, dx = \int \frac{2 \sin x \cos x}{t^2} \left(-\frac{dt}{b \sin x}\right) \]
This simplifies to:
\[ -\frac{2}{b} \int \frac{\cos x}{t^2} \, dt \]

### Step 4: Express \( \cos x \) in terms of \( t \)
From our substitution, we know:
\[ \cos x = \frac{t - a}{b} \]
Thus, we can rewrite the integral as:
\[ -\frac{2}{b} \int \frac{(t - a)/b}{t^2} \, dt = -\frac{2}{b^2} \int \frac{t - a}{t^2} \, dt \]

### Step 5: Split the integral
Now we can split the integral:
\[ -\frac{2}{b^2} \left( \int \frac{1}{t} \, dt - a \int \frac{1}{t^2} \, dt \right) \]

### Step 6: Integrate
The integrals can be computed as follows:
1. \( \int \frac{1}{t} \, dt = \log |t| \)
2. \( \int \frac{1}{t^2} \, dt = -\frac{1}{t} \)

Thus, we have:
\[ -\frac{2}{b^2} \left( \log |t| + a \left(-\frac{1}{t}\right) \right) + C \]
This simplifies to:
\[ -\frac{2}{b^2} \log |t| + \frac{2a}{b^2 t} + C \]

### Step 7: Substitute back for \( t \)
Finally, we substitute back \( t = a + b \cos x \):
\[ -\frac{2}{b^2} \log |a + b \cos x| + \frac{2a}{b^2 (a + b \cos x)} + C \]

### Final Answer
Thus, the final answer is:
\[ -\frac{2}{b^2} \log |a + b \cos x| + \frac{2a}{b^2 (a + b \cos x)} + C \]

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