**Answer **
int 3sin(2x) dx = -(3cos(2x))/2+C
Explanation
You could solve this integral using u -substitution. Typically, we pick u to be our "inner" function, so here, let's denote u=2x which is inside the 3sin(2x)=3sin(u). Now, we can integrate, but we need to make sure all of our variables are in terms of u, so we need to find dx in terms of u.
u=2x
->du=2dx
-> dx=1/2 du
int 3sin(2x) dx= int 3sin(u) *1/2 du=int 3/2 sin(u) du= -3/2cos(u) + C
Since the original question was in terms of x, we want to make sure our final answer is also in terms of x. Remember, u=2x so we just sub back in:
-3/2cos(u) + C=-3/2 cos (2x)+C Which you can rewrite as -(3cos(2x))/2+C if you wanted to.
Alternatively...
You could also solve this integral through inspection, since we easily know the integral of sin (x) alone is supposed to be -cos(x)+C. But we need the integral of 3sin (2x). So we can pull out the constant 3 from the integral
int 3sin(2x) dx=3 int sin(2x) dx
Our lives would be easier if we had 2sin(2x) because that is easily integrated...and we can do that! As long as we multiply by 1/2 on the outside of the integral to cancel it out. So we end up with:
int 3sin(2x) dx=3 int sin(2x) dx=3*1/2int 2sin(2x)dx= -3/2cos(2x)+C
