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.