`int(sin2x)/(sin5x.sin3x)dx=`

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. `3.log(sin3x)-5.log(sin5x)+c`

Answer

To solve the integral \( \int \frac{\sin 2x}{\sin 5x \sin 3x} \, dx \), we can follow these steps:

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

### Step 2: Use the Sine Angle Difference Identity
We can express \(\sin 2x\) using the sine angle difference identity:
\[ \sin 2x = \sin(5x - 3x) = \sin 5x \cos 3x - \cos 5x \sin 3x \]
Thus, we can rewrite the integral as:
\[ I = \int \frac{\sin 5x \cos 3x - \cos 5x \sin 3x}{\sin 5x \sin 3x} \, dx \]

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

### Step 4: Integrate Each Term
Now we integrate each term separately:
1. For \( \int \cot 3x \, dx \):
\[ \int \cot 3x \, dx = \frac{1}{3} \ln |\sin 3x| + C_1 \]
2. For \( \int \cot 5x \, dx \):
\[ \int \cot 5x \, dx = \frac{1}{5} \ln |\sin 5x| + C_2 \]

### Step 5: Combine the Results
Combining the results from the two integrals, we have:
\[ I = \frac{1}{3} \ln |\sin 3x| - \frac{1}{5} \ln |\sin 5x| + C \]
where \( C = C_1 - C_2 \) is the constant of integration.

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{\sin 2x}{\sin 5x \sin 3x} \, dx = \frac{1}{3} \ln |\sin 3x| - \frac{1}{5} \ln |\sin 5x| + C \]