Evaluate :
`int sin x sin 2xdx`

Answer

To evaluate the integral \( \int \sin x \sin 2x \, dx \), we can use the product-to-sum identities or substitution. Here, we will use substitution after rewriting \( \sin 2x \).

### Step-by-Step Solution:

1. **Rewrite \( \sin 2x \)**:
\[ \sin 2x = 2 \sin x \cos x \]
Therefore, we can rewrite the integral:
\[ \int \sin x \sin 2x \, dx = \int \sin x (2 \sin x \cos x) \, dx = \int 2 \sin^2 x \cos x \, dx \]

2. **Factor out the constant**:
Since 2 is a constant, we can factor it out of the integral:
\[ = 2 \int \sin^2 x \cos x \, dx \]

3. **Substitution**:
Let \( t = \sin x \). Then, the derivative \( dt = \cos x \, dx \) or \( \cos x \, dx = dt \). Substituting these into the integral gives:
\[ = 2 \int t^2 \, dt \]

4. **Integrate**:
Now we can integrate \( t^2 \):
\[ = 2 \cdot \frac{t^3}{3} + C = \frac{2}{3} t^3 + C \]

5. **Back Substitute**:
Substitute back \( t = \sin x \):
\[ = \frac{2}{3} \sin^3 x + C \]

6. **Alternative Form**:
If desired, we can express \( \sin^3 x \) in terms of \( \sin 3x \):
\[ \sin 3x = 3 \sin x - 4 \sin^3 x \implies \sin^3 x = \frac{3 \sin x - \sin 3x}{4} \]
Substituting this back gives:
\[ = \frac{2}{3} \left(\frac{3 \sin x - \sin 3x}{4}\right) + C = \frac{1}{2} \sin x - \frac{1}{6} \sin 3x + C \]

### Final Answer:
Thus, the final answer can be expressed as:
\[ \int \sin x \sin 2x \, dx = \frac{2}{3} \sin^3 x + C \quad \text{or} \quad \frac{1}{2} \sin x - \frac{1}{6} \sin 3x + C \]