How do you integrate #int xcosx# by parts from #[0,pi/2]#?

Question

16682 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-21T09:01:30

#int_0^(pi/2)xcosxdx=pi/2-1#

Parts formula.

#intu(dv)/(dx)dx=uv-intv(du)/(dx)dx#

#int_0^(pi/2)xcosxdx#

#u=x=>(du)/(dx)=1#

#(dv)/(dx)=cosx=>v=sinx#

#I=[xsinx-int(sinx)dx]_0^(pi/2)#

#I=[xsinx+cosx]_0^(pi/2)#

now substitute the limits in

#I=[(pi/2)sin(pi/2)+cos(pi/2)]-[0xxsin0+cos0]#

#I=[(pi/2)cancel(sin(pi/2))^1+cancel(cos(pi/2))^0]-[cancel(0xxsin0)^0+cancel(cos0)^1]#

#int_0^(pi/2)xcosxdx=pi/2-1#