antiderivatives?

Question

i can't seem to find the ones for these

3e^2x

3sin2x

Answer 1

I'm assuming that the first one is 3e^(2x) and not 3e^(2)x. If it is the later, you have to do integration by parts.

1. (3/2)e^2x + c
2. (-3/2)cos2x + c

Answer 2

Question 1
I = 3 ∫ e^(2x) dx
Let u = 2x
du = 2 dx
dx = du / 2
I = (3/2) ∫ e^u du
I = (3/2) e^u + C
I = (3/2) e^(2x) + C

Question 2
I = 3 ∫ sin 2x dx
let u = 2x
du = 2 dx
dx = du / 2
I = (3/2) ∫ sin u du
I = - (3/2) cos u + C
I = - (3/2) cos 2x + C

Answer 3

Antiderivatives is the same as Integrals?
2e^2x
let 2x=t
2dx=dt
dx=dt/2
(3/2)integral(e^t)
=(3/2)e^(2x)+c

3integ(sin2x)
(-3/2)cos(2x)+c

Answer 4

Think ahead of what derivatives would lead to these results. The derivative of e^(2x) would be 2e^(2x), so we need to get a 2 in front:

∫ 3e^(2x) dx
3 ∫ e^(2x) dx
(3/2) ∫ 2 e^(2x) dx
(3/2) e^(2x) + C

Likewise, the derivative of cos(2x) is -sin(2x) * 2, so we need to get a -2 in the integral:

∫ 3 sin(2x) dx
3 ∫ sin(2x) dx
-3 ∫ -sin(2x) dx
(-3/2) ∫ -sin(2x) * 2 dx
(-3/2) cos(2x) + C

Answer 5

(3/2)e^2x + C

-(3/2)cos2x +C