Evaluate. `int(dx)/(sinx+sin2x)`

Answer

To evaluate the integral \( I = \int \frac{dx}{\sin x + \sin 2x} \), we will follow these steps:

### Step 1: Rewrite \(\sin 2x\)
We know that \(\sin 2x = 2 \sin x \cos x\). Thus, we can rewrite the integral as:
\[ I = \int \frac{dx}{\sin x + 2 \sin x \cos x} \]

### Step 2: Factor out \(\sin x\)
Now, we can factor \(\sin x\) from the denominator:
\[ I = \int \frac{dx}{\sin x (1 + 2 \cos x)} \]

### Step 3: Rewrite the integral
Next, we can rewrite the integral as:
\[ I = \int \frac{1}{\sin x} \cdot \frac{dx}{1 + 2 \cos x} \]
This can be expressed as:
\[ I = \int \frac{\sin x \, dx}{\sin^2 x (1 + 2 \cos x)} \]

### Step 4: Substitute \(t = \cos x\)
Let \(t = \cos x\), then \(dt = -\sin x \, dx\) or \(dx = -\frac{dt}{\sin x}\). Substituting this into the integral gives:
\[ I = \int \frac{-dt}{(1 - t^2)(1 + 2t)} \]

### Step 5: Perform partial fraction decomposition
We need to perform partial fraction decomposition on:
\[ \frac{1}{(1 - t^2)(1 + 2t)} = \frac{A}{1 - t} + \frac{B}{1 + t} + \frac{C}{1 + 2t} \]
Multiplying through by the denominator gives:
\[ 1 = A(1 + t)(1 + 2t) + B(1 - t)(1 + 2t) + C(1 - t^2) \]

### Step 6: Solve for coefficients \(A\), \(B\), and \(C\)
To find \(A\), \(B\), and \(C\), we can substitute convenient values for \(t\):
1. Set \(t = 1\) to find \(A\).
2. Set \(t = -1\) to find \(B\).
3. Set \(t = -\frac{1}{2}\) to find \(C\).

After solving these equations, we find:
- \(A = \frac{1}{6}\)
- \(B = -\frac{1}{2}\)
- \(C = \frac{4}{3}\)

### Step 7: Substitute back into the integral
Now we can rewrite the integral using the coefficients:
\[ I = \int \left( \frac{1/6}{1 - t} - \frac{1/2}{1 + t} + \frac{4/3}{1 + 2t} \right) dt \]

### Step 8: Integrate each term
Integrating each term separately gives:
\[ I = \frac{1}{6} \ln |1 - t| - \frac{1}{2} \ln |1 + t| + \frac{4}{3} \ln |1 + 2t| + C \]

### Step 9: Substitute back \(t = \cos x\)
Finally, substituting back \(t = \cos x\) gives:
\[ I = \frac{1}{6} \ln |1 - \cos x| - \frac{1}{2} \ln |1 + \cos x| + \frac{4}{3} \ln |1 + 2 \cos x| + C \]

### Final Answer
Thus, the evaluated integral is:
\[ I = \frac{1}{6} \ln |1 - \cos x| - \frac{1}{2} \ln |1 + \cos x| + \frac{4}{3} \ln |1 + 2 \cos x| + C \]