How do you integrate #int xcos(1/2)x# from 0 to pi by integration by parts method?
1 Answer
Explanation:
If this was meant to be:
#intxcos(1/2)dx#
Integration by parts is not necessary, as
#cos(1/2)*intxdx#
#=>cos(1/2)*(x^2)/2#
Evaluating from
#=>cos(1/2)*(pi^2/2-0)#
#=>(pi^2cos(1/2))/2#
If this was meant to be (I assume):
#intxcos(1/2x)dx#
The integration by parts method can be summarized as:
#uv-intvdu#
We pick one of the terms and set it equal to
In this case, I would set
#u=x,du=dx#
#dv=cos(1/2x), v=2sin(1/2x)#
#=>2xsin(x/2)-2intsin(1/2x)dx#
Now we can use substitution to evaluate the integral:
#u=1/2x, du=1/2dx=>2du=dx#
#=>2xsin(x/2)-2(2)intsin(u)du#
Evaluating:
#2xsin(x/2)-4(-cos(u))#
#=>2xsin(x/2)+4cos(u)#
Substituting back in for
#=>2xsin(x/2)+4cos(x/2)#
Now we can evaluate from
#[2(pi)sin(pi/2)+4cos(pi/2)]-[2(0)sin(0/2)+4cos(0/2)]#
#=>[2(pi)sin(pi/2)+4cos(pi/2)]-4cos(0)#
#=>2pi(1)+0-4(1)#
#=>2pi-4#
Hope this helps!