Question

How do you integrate `int 2x sin 4x dx`?

Answer 1

Use integration by parts to find that

`int2xsin(4x)dx = 1/8sin(4x)-1/2xcos(4x)+C`

Explanation:

Using integration by parts:

Let `u = x` and `dv = sin(4x)dx`
Then `du = dx` and `v = -1/4cos(4x)`

So, by the integration by parts formula `intudv = uv - intvdu`

`2intxsin(4x)dx = 2(-1/4xcos(4x)-int((-1/4)cos(4x)dx)`

`=-1/2(xcos(4x)-intcos(4x)dx)`

`= -1/2(xcos(4x)-1/4sin(4x) + C)`

`= 1/8sin(4x)-1/2xcos(4x)+C`