Question

How do you integrate `int xcos(1/2)x` from 0 to pi by integration by parts method?

Answer 1

`intxcos(1/2x)dx=2pi-4`

Explanation:

If this was meant to be:

`intxcos(1/2)dx`

Integration by parts is not necessary, as `cos(1/2)` is a constant. You could bring it outside of the integral:

`cos(1/2)*intxdx`

`=>cos(1/2)*(x^2)/2`

Evaluating from `0->pi`:

`=>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 `u`. This term we will derive. The other term we set equal to `dv`, and this we will "anti-derive." Figuring out which variable to derive and which to anti-derive is usually the more challenging part. With practice it becomes a much quicker process.

In this case, I would set `u=x` and `dv=cos(1/2x)`. This will give `du=dx` and makes one less term for us to have to worry about.

`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 `u`:

`=>2xsin(x/2)+4cos(x/2)`

Now we can evaluate from `0->pi`:

`[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!