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

Answer

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{\sin x}{\sin 3x} \, dx \]

### Step 2: Use the Identity for Sine
Recall the identity for sine:
\[ \sin 3x = 3 \sin x - 4 \sin^3 x \]
Thus, we can rewrite the integral as:
\[ I = \int \frac{\sin x}{3 \sin x - 4 \sin^3 x} \, dx \]

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

### Step 4: Use a Trigonometric Substitution
Let \( u = \sin x \), then \( du = \cos x \, dx \) and \( dx = \frac{du}{\sqrt{1-u^2}} \). The limits of integration are not specified, so we can proceed with indefinite integration:
\[ I = \int \frac{1}{3 - 4u^2} \cdot \frac{du}{\sqrt{1-u^2}} \]

### Step 5: Solve the Integral
This integral can be solved using the method of partial fractions or trigonometric identities. We can express:
\[ \frac{1}{3 - 4u^2} = \frac{1}{\sqrt{3}} \cdot \frac{1}{1 - \frac{4}{3}u^2} \]
This can be integrated using the arctangent function:
\[ I = \frac{1}{\sqrt{3}} \int \frac{1}{1 - \frac{4}{3}u^2} \, du \]
This integral leads to:
\[ I = \frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{\frac{4}{3}}} \cdot \tan^{-1}\left(\sqrt{\frac{4}{3}} u\right) + C \]
Substituting back \( u = \sin x \):
\[ I = \frac{1}{2} \tan^{-1}\left(\frac{2 \sin x}{\sqrt{3}}\right) + C \]

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

---