`int(sin2x)/(sin5xsin3x)dx` is equal to

Options

A. `log sin3x-log sin5x+c`
B. `(1)/(3)log sin3x+(1)/(5)log sin5x+c`
C. `(1)/(3)log sin3x-(1)/(5)log sin5x+c`
D. `3log sin3x-5log sin5x+c`

Explanation

To solve the integral \( \int \frac{\sin 2x}{\sin 5x \sin 3x} \, dx \), we can use the identity for the sine of a difference. Here are the steps to solve it:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{\sin 2x}{\sin 5x \sin 3x} \, dx \]

### Step 2: Use the Sine Difference Identity
We can express \( \sin 2x \) in terms of \( \sin(5x - 3x) \):
\[ \sin 2x = \sin(5x - 3x) = \sin 5x \cos 3x - \cos 5x \sin 3x \]
Thus, we rewrite the integral:
\[ I = \int \frac{\sin 5x \cos 3x - \cos 5x \sin 3x}{\sin 5x \sin 3x} \, dx \]

### Step 3: Split the Integral
Now we can split the integral into two parts:
\[ I = \int \left( \frac{\sin 5x \cos 3x}{\sin 5x \sin 3x} - \frac{\cos 5x \sin 3x}{\sin 5x \sin 3x} \right) \, dx \]
This simplifies to:
\[ I = \int \left( \cot 3x - \cot 5x \right) \, dx \]

### Step 4: Integrate Each Term
Now we can integrate each term separately:
\[ I = \int \cot 3x \, dx - \int \cot 5x \, dx \]

Using the integral formula \( \int \cot kx \, dx = \frac{1}{k} \ln |\sin kx| + C \), we get:
\[ I = \frac{1}{3} \ln |\sin 3x| - \frac{1}{5} \ln |\sin 5x| + C \]

### Final Result
Thus, the final result for the integral is:
\[ I = \frac{1}{3} \ln |\sin 3x| - \frac{1}{5} \ln |\sin 5x| + C \]