`int sin 3x dx`

Answer

To solve the integral \( \int \sin(3x) \, dx \), we can use a substitution method. Here's a step-by-step solution:

### Step 1: Substitution
Let \( t = 3x \). Then, we differentiate both sides with respect to \( x \):
\[ dt = 3 \, dx \quad \Rightarrow \quad dx = \frac{dt}{3} \]

### Step 2: Rewrite the Integral
Now, substitute \( t \) and \( dx \) into the integral:
\[ \int \sin(3x) \, dx = \int \sin(t) \cdot \frac{dt}{3} \]

### Step 3: Factor Out the Constant
We can factor out the constant \( \frac{1}{3} \):
\[ \int \sin(3x) \, dx = \frac{1}{3} \int \sin(t) \, dt \]

### Step 4: Integrate
Now, we know the integral of \( \sin(t) \):
\[ \int \sin(t) \, dt = -\cos(t) + C \]
where \( C \) is the constant of integration.

### Step 5: Substitute Back
Now, substitute back \( t = 3x \):
\[ \frac{1}{3} \int \sin(t) \, dt = \frac{1}{3} \left(-\cos(t) + C\right) = -\frac{1}{3} \cos(3x) + C \]

### Final Answer
Thus, the integral \( \int \sin(3x) \, dx \) is:
\[ -\frac{1}{3} \cos(3x) + C \]

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