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.