`int sinx/sin(3x) dx=`

Answer

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

### Step 1: Use the identity for \(\sin(3x)\)
We know that:
\[ \sin(3x) = 3\sin x - 4\sin^3 x \]
Thus, we can rewrite the integral:
\[ \int \frac{\sin x}{\sin(3x)} \, dx = \int \frac{\sin x}{3\sin x - 4\sin^3 x} \, dx \]

**Hint for Step 1:** Recall the trigonometric identity for \(\sin(3x)\) to simplify the integral.

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

**Hint for Step 2:** Look for common factors in the numerator and denominator to simplify the expression.

### Step 3: Rewrite the integral in terms of \(\tan x\)
Next, we can use the substitution \(\sin^2 x = \frac{1}{1 + \tan^2 x}\) and rewrite the integral:
\[ \int \frac{1}{3 - 4\sin^2 x} \, dx = \int \frac{1}{3 - 4\left(\frac{1}{1 + \tan^2 x}\right)} \, dx \]
This simplifies to:
\[ \int \frac{1 + \tan^2 x}{3(1 + \tan^2 x) - 4} \, dx = \int \frac{1 + \tan^2 x}{3 + 3\tan^2 x - 4} \, dx = \int \frac{1 + \tan^2 x}{-1 + 3\tan^2 x} \, dx \]

**Hint for Step 3:** Use the substitution \(t = \tan x\) to simplify the integral further.

### Step 4: Substitute \(t = \tan x\)
Now, we differentiate \(t = \tan x\):
\[ dt = \sec^2 x \, dx \quad \text{or} \quad dx = \frac{dt}{\sec^2 x} = \frac{dt}{1 + t^2} \]
Substituting this into the integral gives:
\[ \int \frac{1 + t^2}{-1 + 3t^2} \cdot \frac{dt}{1 + t^2} = \int \frac{dt}{-1 + 3t^2} \]

**Hint for Step 4:** Remember to change the differential when you substitute \(t = \tan x\).

### Step 5: Integrate using the formula for \(\int \frac{1}{a^2 - x^2} \, dx\)
Now, we can rewrite the integral:
\[ \int \frac{dt}{-1 + 3t^2} = \int \frac{dt}{\sqrt{3^2 - t^2}} \quad \text{(after factoring out -1)} \]
Using the formula \(\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln \left| \frac{a + x}{a - x} \right| + C\), we have:
\[ \frac{1}{2\sqrt{3}} \ln \left| \frac{\sqrt{3} + t}{\sqrt{3} - t} \right| + C \]

**Hint for Step 5:** Familiarize yourself with the standard integral formulas for logarithmic functions.

### Step 6: Substitute back \(t = \tan x\)
Finally, we substitute back \(t = \tan x\):
\[ \frac{1}{2\sqrt{3}} \ln \left| \frac{\sqrt{3} + \tan x}{\sqrt{3} - \tan x} \right| + C \]

**Hint for Step 6:** Always remember to revert back to the original variable after integration.

### Final Result
Thus, the final answer for the integral is:
\[ \int \frac{\sin x}{\sin(3x)} \, dx = \frac{1}{2\sqrt{3}} \ln \left| \frac{\sqrt{3} + \tan x}{\sqrt{3} - \tan x} \right| + C \]