Solve `intsin2x.sin3xdx`

Explanation

To solve the integral \( \int \sin 2x \sin 3x \, dx \), we can use a trigonometric identity to simplify the expression before integrating. Here’s a step-by-step solution:

### Step 1: Use the trigonometric identity
We know from trigonometric identities that:
\[ 2 \sin A \sin B = \cos(A - B) - \cos(A + B) \]
In our case, let \( A = 2x \) and \( B = 3x \). Therefore, we can write:
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