To evaluate the integral \(\int_{0}^{\frac{\pi}{4}} \sin(2x) \, dx\), we can follow these steps:

1. **Substitute**:
   Let \(u = 2x\). Then, \(du = 2 \, dx\), or equivalently, \(dx = \frac{du}{2}\).

2. **Change the limits of integration**:
   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**:
   The integral becomes:
   \[
   \int_{0}^{\frac{\pi}{4}} \sin(2x) \, dx = \int_{0}^{\frac{\pi}{2}} \sin(u) \cdot \frac{1}{2} \, du = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin(u) \, du
   \]

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

5. **Apply the limits**:
   \[
   \frac{1}{2} \left[ -\cos\left(\frac{\pi}{2}\right) + \cos(0) \right] = \frac{1}{2} \left[ -0 + 1 \right] = \frac{1}{2} \cdot 1 = \frac{1}{2}
   \]

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

Would you like more details on this solution or have any questions?

Here are some related questions for further practice:
1. Evaluate \(\int_{0}^{\frac{\pi}{2}} \cos(3x) \, dx\).
2. Evaluate \(\int_{0}^{\pi} \sin(4x) \, dx\).
3. Find the antiderivative of \(\sin(kx)\) for a constant \(k\).
4. Evaluate \(\int_{0}^{\frac{\pi}{3}} \sin(2x) \, dx\).
5. Find \(\int_{0}^{\frac{\pi}{6}} \sin(5x) \, dx\).
6. Compute \(\int_{0}^{1} \sin(2\pi x) \, dx\).
7. Determine \(\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin(2x) \, dx\).
8. Evaluate \(\int_{0}^{\frac{\pi}{4}} \cos(2x) \, dx\).

**Tip**: When dealing with integrals involving trigonometric functions, substitution can often simplify the process.

Learn how to evaluate the integral ∫ from 0 to π/4 sin(2x) dx with a detailed step-by-step solution using trigonometric substitution. This problem involves integral calculus and trigonometric methods. Suitable for advanced high school students and those preparing for calculus exams.

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