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 #
