Evaluate `int (sin 2x)/((1+sin x)(2+sin x))dx`

Answer

To evaluate the integral

\[ I = \int \frac{\sin 2x}{(1 + \sin x)(2 + \sin x)} \, dx, \]

we can follow these steps:

### Step 1: Rewrite the Integral
We know that \(\sin 2x = 2 \sin x \cos x\). Therefore, we can rewrite the integral as:

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

### Step 2: Substitution
Let \(t = \sin x\). Then, \(dt = \cos x \, dx\), which means \(dx = \frac{dt}{\cos x}\). We can also express \(\cos x\) in terms of \(t\) using the identity \(\cos^2 x = 1 - \sin^2 x\), so \(\cos x = \sqrt{1 - t^2}\).

Substituting these into the integral gives:

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

### Step 3: Partial Fraction Decomposition
Next, we will perform partial fraction decomposition on the integrand:

\[ \frac{2t}{(1 + t)(2 + t)} = \frac{A}{1 + t} + \frac{B}{2 + t}. \]

Multiplying through by the denominator \((1 + t)(2 + t)\) gives:

\[ 2t = A(2 + t) + B(1 + t). \]

### Step 4: Solve for A and B
Expanding the right side:

\[ 2t = (2A + B) + (A + B)t. \]

By comparing coefficients, we have:

1. \(A + B = 2\) (coefficient of \(t\))
2. \(2A + B = 0\) (constant term)

From the first equation, we can express \(B\) in terms of \(A\):

\[ B = 2 - A. \]

Substituting into the second equation:

\[ 2A + (2 - A) = 0 \implies A + 2 = 0 \implies A = -2. \]

Then substituting back to find \(B\):

\[ B = 2 - (-2) = 4. \]

Thus, we have:

\[ \frac{2t}{(1 + t)(2 + t)} = \frac{-2}{1 + t} + \frac{4}{2 + t}. \]

### Step 5: Substitute Back into the Integral
Now we can rewrite the integral:

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

### Step 6: Integrate
Now we can integrate term by term:

1. The integral of \(\frac{-2}{1 + t}\) is \(-2 \ln |1 + t|\).
2. The integral of \(\frac{4}{2 + t}\) is \(4 \ln |2 + t|\).

Thus, we have:

\[ I = -2 \ln |1 + t| + 4 \ln |2 + t| + C. \]

### Step 7: Substitute Back for \(t\)
Since \(t = \sin x\), we substitute back to get:

\[ I = 4 \ln |2 + \sin x| - 2 \ln |1 + \sin x| + C. \]

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

\[ \int \frac{\sin 2x}{(1 + \sin x)(2 + \sin x)} \, dx = 4 \ln |2 + \sin x| - 2 \ln |1 + \sin x| + C. \]

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