Let $\alpha,\beta$ be the distinct positive roots of the equation of $2x=\tan x$.
Evaluate

\[\int_0^1 \sin \alpha x\sin \beta x\ dx\]
1. We start with the integral \(\int_0^1 \sin \alpha x \sin \beta x \, dx\). Using the product-to-sum identities, we can rewrite the integrand:
   \[
   \sin \alpha x \sin \beta x = \frac{1}{2} [\cos (\alpha - \beta)x - \cos (\alpha + \beta)x]
   \]
   Therefore, the integral becomes:
   \[
   \int_0^1 \sin \alpha x \sin \beta x \, dx = \frac{1}{2} \int_0^1 [\cos (\alpha - \beta)x - \cos (\alpha + \beta)x] \, dx
   \]

2. We can split the integral into two separate integrals:
   \[
   \int_0^1 \sin \alpha x \sin \beta x \, dx = \frac{1}{2} \left( \int_0^1 \cos (\alpha - \beta)x \, dx - \int_0^1 \cos (\alpha + \beta)x \, dx \right)
   \]

3. We now evaluate each integral separately. For the first integral:
   \[
   \int_0^1 \cos (\alpha - \beta)x \, dx
   \]
   Using the antiderivative of \(\cos kx\), we get:
   \[
   \int \cos (\alpha - \beta)x \, dx = \frac{\sin (\alpha - \beta)x}{\alpha - \beta}
   \]
   Evaluating from 0 to 1:
   \[
   \left[ \frac{\sin (\alpha - \beta)x}{\alpha - \beta} \right]_0^1 = \frac{\sin (\alpha - \beta)}{\alpha - \beta}
   \]

4. For the second integral:
   \[
   \int_0^1 \cos (\alpha + \beta)x \, dx
   \]
   Using the antiderivative of \(\cos kx\), we get:
   \[
   \int \cos (\alpha + \beta)x \, dx = \frac{\sin (\alpha + \beta)x}{\alpha + \beta}
   \]
   Evaluating from 0 to 1:
   \[
   \left[ \frac{\sin (\alpha + \beta)x}{\alpha + \beta} \right]_0^1 = \frac{\sin (\alpha + \beta)}{\alpha + \beta}
   \]

5. Combining the results, we have:
   \[
   \int_0^1 \sin \alpha x \sin \beta x \, dx = \frac{1}{2} \left( \frac{\sin (\alpha - \beta)}{\alpha - \beta} - \frac{\sin (\alpha + \beta)}{\alpha + \beta} \right)
   \]

6. Given that \(\alpha\) and \(\beta\) are distinct positive roots of the equation \(2x = \tan x\), we have:
   \[
   2\alpha = \tan \alpha \quad \text{and} \quad 2\beta = \tan \beta
   \]

7. Using the trigonometric identity for the tangent function, we can express:
   \[
   2(\alpha - \beta) = \frac{\sin (\alpha - \beta)}{\cos \alpha \cos \beta} \quad \Rightarrow \quad \frac{\sin (\alpha - \beta)}{2(\alpha - \beta)} = \cos \alpha \cos \beta
   \]
   Similarly,
   \[
   2(\alpha + \beta) = \frac{\sin (\alpha + \beta)}{\cos \alpha \cos \beta} \quad \Rightarrow \quad \frac{\sin (\alpha + \beta)}{2(\alpha + \beta)} = \cos \alpha \cos \beta
   \]

8. Substituting these results back into our integral expression, we get:
   \[
   \int_0^1 \sin \alpha x \sin \beta x \, dx = \frac{1}{2} \left( 2 \cos \alpha \cos \beta - 2 \cos \alpha \cos \beta \right) = 0
   \]

Therefore, the integral evaluates to zero.

The final answer is \(\boxed{0}\).