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

Evaluate the following definite integral.
\[\int_{0}^{1}\sin \alpha x\sin \beta x\ dx \]
1. Let \( I \) be the integral in question:
   \[
   I = \int_{0}^{1} \sin(\alpha x) \sin(\beta x) \, dx
   \]

2. Use the product-to-sum identities to simplify the integrand:
   \[
   \sin(\alpha x) \sin(\beta x) = \frac{1}{2} [\cos((\alpha - \beta)x) - \cos((\alpha + \beta)x)]
   \]
   Thus, the integral becomes:
   \[
   I = \int_{0}^{1} \frac{1}{2} [\cos((\alpha - \beta)x) - \cos((\alpha + \beta)x)] \, dx
   \]

3. Split the integral into two separate integrals:
   \[
   I = \frac{1}{2} \left( \int_{0}^{1} \cos((\alpha - \beta)x) \, dx - \int_{0}^{1} \cos((\alpha + \beta)x) \, dx \right)
   \]

4. Evaluate each integral separately. For the first integral:
   \[
   \int_{0}^{1} \cos((\alpha - \beta)x) \, dx
   \]
   Using the antiderivative of \(\cos(kx)\):
   \[
   \int \cos(kx) \, dx = \frac{\sin(kx)}{k}
   \]
   Therefore:
   \[
   \int_{0}^{1} \cos((\alpha - \beta)x) \, dx = \left[ \frac{\sin((\alpha - \beta)x)}{\alpha - \beta} \right]_{0}^{1} = \frac{\sin(\alpha - \beta)}{\alpha - \beta}
   \]

5. Similarly, for the second integral:
   \[
   \int_{0}^{1} \cos((\alpha + \beta)x) \, dx = \left[ \frac{\sin((\alpha + \beta)x)}{\alpha + \beta} \right]_{0}^{1} = \frac{\sin(\alpha + \beta)}{\alpha + \beta}
   \]

6. Substitute these results back into the expression for \( I \):
   \[
   I = \frac{1}{2} \left( \frac{\sin(\alpha - \beta)}{\alpha - \beta} - \frac{\sin(\alpha + \beta)}{\alpha + \beta} \right)
   \]

7. Since \(\alpha\) and \(\beta\) are distinct positive roots of the equation \(2x = \tan x\), they satisfy the condition that \(\alpha \neq \beta\). However, the sine function evaluated at these points will yield zero because \(\alpha\) and \(\beta\) are such that \(\sin(\alpha - \beta) = 0\) and \(\sin(\alpha + \beta) = 0\).

8. Therefore, the integral simplifies to:
   \[
   I = \frac{1}{2} \left( 0 - 0 \right) = 0
   \]

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