3.) a.) $$\int \sin(2x) dx = $$

Let's solve the integral: $$\int \sin(2x) dx$$ We know that the integral of $$\sin(ax)$$ is $$-\frac{1}{a}\cos(ax) + C$$. Therefore, $$\int \sin(2x) dx = -\frac{1}{2} \cos(2x) + C$$ So, the result is: $$-\frac{1}{2} \cos(2x) + C$$ Answer: $$-\frac{1}{2} \cos(2x) + C$$