# 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!