**Answer **
int 3sin(2x) dx = -(3cos(2x))/2+C∫3sin(2x)dx=−3cos(2x)2+C
Explanation
You could solve this integral using u u -substitution. Typically, we pick uu to be our "inner" function, so here, let's denote u=2xu=2x which is inside the 3sin(2x)=3sin(u)3sin(2x)=3sin(u). Now, we can integrate, but we need to make sure all of our variables are in terms of uu, so we need to find dxdx in terms of uu.
u=2xu=2x
->→du=2dxdu=2dx
->→ dx=1/2 dudx=12du
int 3sin(2x) dx= int 3sin(u) *1/2 du=int 3/2 sin(u) du= -3/2cos(u) + C∫3sin(2x)dx=∫3sin(u)⋅12du=∫32sin(u)du=−32cos(u)+C
Since the original question was in terms of xx, we want to make sure our final answer is also in terms of xx. Remember, u=2xu=2x so we just sub back in:
-3/2cos(u) + C=-3/2 cos (2x)+C−32cos(u)+C=−32cos(2x)+C Which you can rewrite as -(3cos(2x))/2+C−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)sin(x) alone is supposed to be -cos(x)+C−cos(x)+C. But we need the integral of 3sin (2x)3sin(2x). So we can pull out the constant 3 from the integral
int 3sin(2x) dx=3 int sin(2x) dx∫3sin(2x)dx=3∫sin(2x)dx
Our lives would be easier if we had 2sin(2x)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 ∫3sin(2x)dx=3∫sin(2x)dx=3⋅12∫2sin(2x)dx=−32cos(2x)+C
