Question

How do you integrate `intx cos 2x dx`?

Answer 1

`intxcos(2x)dx=` by parts:
`=xsin(2x)/2-[intsin(2x)/2dx]=`
`=xsin(2x)/2+1/4cos(2x)+c`

Answer 2

Use integration by parts

`\intu \quad dv=uv-intv du`

Let `u=x`, `\quad \implies du=dx`

and let `\quad \quaddv=cos(2x)dx`, `\implies v=1/2sin(2x)`

Now integrate by parts

`intxcos(2x)dx=intu\quaddv=uv-intvdu`

`\quad \quad \quad \quad \quad \quad \quad \quad\quad \quad \quad \quad\quad \quad \quad \quad\quad \quad \quad \quad =x*1/2sin(2x)-int1/2sin(2x)dx`

`\quad \quad \quad \quad \quad \quad \quad \quad\quad \quad \quad \quad\quad \quad \quad \quad\quad \quad \quad \quad =x/2sin(2x)+1/4cos(2x)+C`

where `C` is the constant of integration.

Quick note on how to get integration by parts formula:

The differential of `uv` is

`d[uv]=udv+vdu`

`udv=d[uv]-vdu`

Integrate both sides

`\int udv=int d[uv]-int vdu`

`\int u dv=uv-intvdu`