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To solve the integral \(\int_{0}^{\pi/4} \sin(2x) \, dx\), we will use the following steps:

1. **Substitution**:
   Let \( u = 2x \). Then \( du = 2 \, dx \), which implies \( dx = \frac{1}{2} \, du \).

2. **Changing the Limits**:
   When \( x = 0 \):
   \[
   u = 2 \cdot 0 = 0
   \]

   When \( x = \frac{\pi}{4} \):
   \[
   u = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2}
   \]

3. **Rewrite the Integral**:
   Substitute \( u \) and \( dx \) into the integral:
   \[
   \int_{0}^{\pi/4} \sin(2x) \, dx = \int_{0}^{\pi/2} \sin(u) \cdot \frac{1}{2} \, du = \frac{1}{2} \int_{0}^{\pi/2} \sin(u) \, du
   \]

4. **Evaluate the Integral**:
   The integral of \(\sin(u)\) is \(-\cos(u)\):
   \[
   \frac{1}{2} \int_{0}^{\pi/2} \sin(u) \, du = \frac{1}{2} \left[ -\cos(u) \right]_{0}^{\pi/2}
   \]

5. **Apply the Limits**:
   Evaluate \(-\cos(u)\) from \(0\) to \(\pi/2\):
   \[
   \frac{1}{2} \left[ -\cos\left( \frac{\pi}{2} \right) + \cos(0) \right] = \frac{1}{2} \left[ -0 + 1 \right] = \frac{1}{2} \left[ 1 \right] = \frac{1}{2}
   \]

So, the value of the integral \(\int_{0}^{\pi/4} \sin(2x) \, dx\) is \(\boxed{\frac{1}{2}}\).

Would you like more details or have any questions about this solution? Here are some related questions for further practice:

1. Evaluate \(\int_{0}^{\pi/6} \cos(3x) \, dx\).
2. Find the antiderivative of \(\sin(5x)\).
3. Determine \(\int_{0}^{\pi/2} \sin(x) \, dx\).
4. Calculate \(\int_{0}^{\pi} \cos(2x) \, dx\).
5. Solve \(\int_{0}^{\pi/4} \sin(x) \, dx\).
6. Evaluate \(\int_{0}^{\pi/3} \cos(4x) \, dx\).
7. Find \(\int_{0}^{2\pi} \sin(x) \, dx\).
8. Determine \(\int_{0}^{\pi/6} \sin(2x) \, dx\).

**Tip**: When dealing with trigonometric integrals, using substitution can simplify the process significantly. Always remember to change the limits of integration when you make a substitution.

Learn how to evaluate the integral ∫(0 to π/4) sin(2x) dx with detailed step-by-step solutions. This problem involves using substitution and integrating trigonometric functions. Suitable for advanced high school students and above.

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