Question

What is `int xsin(x/2-pi)`?

Answer 1

`2xcos(x/2)-4sin(x/2)+C`

Explanation:

First, we can simplify `sin(x/2-pi)` using the sine angle subtraction formula:

`sin(A-B)=sinAcosB-cosAsinB`

Thus,

`sin(x/2-pi)=sin(x/2)cos(pi)-cos(x/2)sin(pi)`

`=sin(x/2)(-1)-cos(x/2)(0)`

`=-sin(x/2)`

Thus the integral is slightly simplified to become

`=-intxsin(x/2)dx`

First, set `t=x/2`. This implies that `dt=1/2dx` and `x=2t`. We will want to multiply the interior of the integral by `1/2` so that we have `1/2dx=t` and balance this by multiplying the exterior by `2`.

`=-2intxsin(x/2)(1/2)dx`

Substitute in `x=2t`, `x/2=t`, and `1/2dx=dt`.

`=-2int2tsin(t)dt=-4inttsin(t)dt`

Here, use integration by parts, which takes the form

`intudv=uv-intvdu`

For `inttsin(t)dt`, set `u=t` and `dv=sin(t)dt`, which imply that `du=dt` and `v=-cos(t)`.

Hence,

`inttsin(t)dt=t(-cos(t))-int(-cos(t))dt`

`=-tcos(t)+sin(t)+C`

Multiply these both by `-4` to see that:

`-4inttsin(t)dt=4tcos(t)-4sin(t)+C`

Now, recall that `t=x/2`:

`intxsin(x/2-pi)dx=4(x/2)cos(x/2)-4sin(x/2)+C`

`=2xcos(x/2)-4sin(x/2)+C`