How do you integrate intx cos 2x dxxcos2xdx?

2 Answers
Jun 5, 2015

intxcos(2x)dx=xcos(2x)dx= by parts:
=xsin(2x)/2-[intsin(2x)/2dx]==xsin(2x)2[sin(2x)2dx]=
=xsin(2x)/2+1/4cos(2x)+c=xsin(2x)2+14cos(2x)+c

Jun 5, 2015

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