To find the indefinite integral of $\sin(2x)$, we proceed as follows:

$\int \sin(2x) \, dx$

1. Use a substitution to simplify the integral. Let $u = 2x$, then $du = 2 \, dx$ or $dx = \frac{du}{2}$.

2. Substitute into the integral:

   $\int \sin(2x) \, dx = \int \sin(u) \cdot \frac{du}{2} = \frac{1}{2} \int \sin(u) \, du$

3. Now integrate $\sin(u)$:

   $\frac{1}{2} \int \sin(u) \, du = \frac{1}{2}(-\cos(u)) = -\frac{1}{2} \cos(u)$

4. Substitute $u = 2x$ back into the result:

   $-\frac{1}{2} \cos(2x) + C$

Thus, the answer is:

$\int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) + C$

Do you want more details on the substitution method or have any other questions?

Related Questions

1. What is the indefinite integral of $\cos(2x)$?
2. How do you find the indefinite integral of $\sin(kx)$ for any constant $k$?
3. What is the integral of $\sin^2(2x)$?
4. How would you solve $\int \cos^2(2x) \, dx$?
5. What is the definite integral of $\sin(2x)$ from $0$ to $\pi$?

Tip: Remember that substitution is a powerful technique for integrals involving functions of $ax + b$, where $a$ is a constant.