Interval find 2x cosine 2xdx?

Answer 1

Integral (2x cos(2x) ) dx

First thing you do is pull the constant out of the integral.

2 * Integral (x cos(2x) ) dx

Now, we solve this by parts.

Let u = x. dv = cos(2x) dx
du = dx. v = (1/2) sin(2x)

Recall that the procedures for doing this by parts goes
uv - Integral (v du), so we have

2 * [x(1/2)sin(2x) - Integral [ (1/2)sin(2x) ] dx ]

If we expand this, we get

xsin(2x) - 2 * Integral [ (1/2)sin(2x) ] dx

Pull the (1/2) out of the other integral; merged with the existing 2 on the outside, should mean they cancel each other out.

xsin(2x) - Integral (sin(2x)) dx

Now we solve as normal.

xsin(2x) - (1/2)(-cos(2x)) + C
xsin(2x) + (1/2)cos(2x) + C

Answer 2

By parts:

u = 2x
du = 2 dx

dv = cos(2x) dx
v = 1/2 sin(2x) dx

∫ u dv = uv - ∫ v du

∫ 2x cos(2x) dx = 1/2 (2x)sin(2x) - 1/2 ∫sin(2x) (2 dx)

= x sin(2x) - ∫sin(2x) dx

= x sin(2x) - (-1/2 cos(2x)) + C

= x sin(2x) + 1/2 cos(2x) + C