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

Explanation

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

### Step 1: Simplify the Integral
We start by rewriting \( \sin 2x \) using the double angle formula:
\[ \sin 2x = 2 \sin x \cos x \]
Thus, the integral becomes:
\[ \int \frac{dx}{\sin x + 2 \sin x \cos x} = \int \frac{dx}{\sin x (1 + 2 \cos x)} \]

### Step 2: Factor Out \( \sin x \)
We can rewrite the integral as:
\[ \int \frac{dx}{\sin x (1 + 2 \cos x)} = \int \frac{1}{\sin x} \cdot \frac{1}{1 + 2 \cos x} \, dx \]

### Step 3: Substitute \( t = \cos x \)
Let \( t = \cos x \), then \( dt = -\sin x \, dx \) or \( dx = -\frac{dt}{\sin x} \). Thus, we rewrite the integral:
\[ \int \frac{1}{\sin x (1 + 2t)} (-dt) = -\int \frac{dt}{1 + 2t} \]

### Step 4: Integrate
Now we can integrate:
\[ -\int \frac{dt}{1 + 2t} = -\frac{1}{2} \ln |1 + 2t| + C \]

### Step 5: Substitute Back
Substituting back \( t = \cos x \):
\[ -\frac{1}{2} \ln |1 + 2 \cos x| + C \]

### Final Answer
Thus, the integral \( \int \frac{dx}{\sin x + \sin 2x} \) evaluates to:
\[ -\frac{1}{2} \ln |1 + 2 \cos x| + C \]

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