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

Question

1716 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-05T02:04:09

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

Answer 2 · 2015-06-05T02:11:17

#\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#