`int (sin 3x)/sin xdx=.....`

Options

A. x-sin 2x+c
B. x+sin 2x+c
C. x+cos 2x+c
D. x-cos 2x+c

Answer

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

### Step 1: Use the identity for \(\sin 3x\)
We start by using the trigonometric identity for \(\sin 3x\):
\[ \sin 3x = 3 \sin x - 4 \sin^3 x \]
Substituting this into the integral gives:
\[ \int \frac{\sin 3x}{\sin x} \, dx = \int \frac{3 \sin x - 4 \sin^3 x}{\sin x} \, dx \]

### Step 2: Simplify the expression
We can simplify the fraction:
\[ \int \frac{3 \sin x - 4 \sin^3 x}{\sin x} \, dx = \int \left(3 - 4 \sin^2 x\right) \, dx \]

### Step 3: Substitute \(\sin^2 x\)
We know that:
\[ \sin^2 x = \frac{1 - \cos 2x}{2} \]
Substituting this into the integral gives:
\[ \int \left(3 - 4 \cdot \frac{1 - \cos 2x}{2}\right) \, dx = \int \left(3 - 2(1 - \cos 2x)\right) \, dx \]

### Step 4: Expand the expression
Now, we expand the expression:
\[ \int \left(3 - 2 + 2 \cos 2x\right) \, dx = \int (1 + 2 \cos 2x) \, dx \]

### Step 5: Integrate term by term
Now we can integrate term by term:
\[ \int 1 \, dx + \int 2 \cos 2x \, dx \]
The first integral gives:
\[ x \]
The second integral can be computed using the formula for the integral of cosine:
\[ \int \cos kx \, dx = \frac{1}{k} \sin kx \]
Thus:
\[ \int 2 \cos 2x \, dx = 2 \cdot \frac{1}{2} \sin 2x = \sin 2x \]

### Step 6: Combine the results
Combining the results from the integrals, we have:
\[ x + \sin 2x + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{\sin 3x}{\sin x} \, dx = x + \sin 2x + C \]