Evaluate `int (sin x)/(sin 3x) dx`

Answer

To evaluate the integral \( \int \frac{\sin x}{\sin 3x} \, dx \), we can follow these steps:

### Step 1: Rewrite \( \sin 3x \)
Using the triple angle formula for sine, we have:
\[ \sin 3x = 3 \sin x - 4 \sin^3 x \]
Thus, we can rewrite the integral as:
\[ \int \frac{\sin x}{3 \sin x - 4 \sin^3 x} \, dx \]

### Step 2: Factor out \( \sin x \)
We can factor \( \sin x \) from the denominator:
\[ \int \frac{\sin x}{\sin x (3 - 4 \sin^2 x)} \, dx = \int \frac{1}{3 - 4 \sin^2 x} \, dx \]

### Step 3: Use a substitution
Let \( t = \tan x \). Then, we know:
\[ \sin^2 x = \frac{t^2}{1+t^2} \quad \text{and} \quad dx = \frac{dt}{1+t^2} \]
Substituting these into the integral gives:
\[ \int \frac{1}{3 - 4 \frac{t^2}{1+t^2}} \cdot \frac{dt}{1+t^2} \]

### Step 4: Simplify the integral
This simplifies to:
\[ \int \frac{1+t^2}{3(1+t^2) - 4t^2} \, dt = \int \frac{1+t^2}{3 - t^2} \, dt \]

### Step 5: Split the integral
We can split the integral into two parts:
\[ \int \frac{1}{3 - t^2} \, dt + \int \frac{t^2}{3 - t^2} \, dt \]

### Step 6: Solve the first integral
The first integral can be solved using the formula for the integral of a rational function:
\[ \int \frac{1}{a^2 - x^2} \, dx = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| + C \]
For \( a = \sqrt{3} \):
\[ \int \frac{1}{3 - t^2} \, dt = \frac{1}{2\sqrt{3}} \ln \left| \frac{\sqrt{3}+t}{\sqrt{3}-t} \right| + C \]

### Step 7: Solve the second integral
For the second integral, we can rewrite it as:
\[ \int \frac{t^2}{3 - t^2} \, dt = \int \left( \frac{3 - (3 - t^2)}{3 - t^2} \right) dt = \int 1 \, dt - \int \frac{3}{3 - t^2} \, dt \]
The first part integrates to \( t \), and the second part can be solved similarly as above.

### Step 8: Combine the results
Combining both integrals gives us the final result in terms of \( t = \tan x \).

### Final Answer
Thus, the evaluated integral is:
\[ \int \frac{\sin x}{\sin 3x} \, dx = \frac{1}{2\sqrt{3}} \ln \left| \frac{\sqrt{3} + \tan x}{\sqrt{3} - \tan x} \right| + C + \tan x - 3 \int \frac{1}{3 - t^2} \, dt \]