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

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

So we can rewrite the integral as:

\[ I = \int \frac{dx}{\sin x + 2 \sin x \cos x} \]

### Step 2: Factor out \(\sin x\)
Factoring out \(\sin x\) from the denominator gives us:

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

### Step 3: Multiply numerator and denominator by \(\sin x\)
Now, we multiply the numerator and denominator by \(\sin x\):

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

### Step 4: Substitute \(\cos x = p\)
Let \(p = \cos x\). Then, \(dx = -\frac{1}{\sin x} dp\). Thus, we can rewrite the integral:

\[ I = \int \frac{-dp}{(1 - p^2)(1 + 2p)} \]

### Step 5: Use partial fractions
We will express the integrand using partial fractions:

\[ \frac{1}{(1 - p^2)(1 + 2p)} = \frac{A}{1 - p} + \frac{B}{1 + p} + \frac{C}{1 + 2p} \]

Multiplying through by the denominator gives:

\[ 1 = A(1 + p)(1 + 2p) + B(1 - p)(1 + 2p) + C(1 - p^2) \]

### Step 6: Solve for coefficients
To find \(A\), \(B\), and \(C\), we can substitute convenient values for \(p\) (like \(p = 1\), \(p = -1\), and \(p = -\frac{1}{2}\)) and solve the resulting equations.

### Step 7: Integrate each term
Once we have \(A\), \(B\), and \(C\), we can integrate each term separately:

\[ I = \int \left( \frac{A}{1 - p} + \frac{B}{1 + p} + \frac{C}{1 + 2p} \right) dp \]

### Step 8: Substitute back
After integrating, we substitute back \(p = \cos x\) to express the integral in terms of \(x\).

### Final Result
The final result will be:

\[ I = \frac{1}{6} \ln |1 - \cos x| + \frac{1}{2} \ln |1 + \cos x| - \frac{2}{3} \ln |1 + 2 \cos x| + C \]

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

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