Question

How do I evaluate #int_(1/6)^(1/2)[csc(pit)][cot(pit)] dt#?

Answer 1

For any integral, first try straightaway integration, then substitution. Do I know a function whose derivative is or involves csc time cot?

Yes, of course: `d/(dx)(cscx)=-cscxcotx`.
That's not quite what we're integrating, but we can fix that with a substitution:
Let `u=pit`. That makes `du=pi dt` and `dt = (du)/pi`
When `x=1/6` (the lower limit of integration), then `u=4*1/6=pi/6`.
When `x=1/2` (the upper limit). then `u=pi/2`. So, substituting, our integral becomes:

`int_(pi/6)^(pi/2)cscucotu (du)/pi=1/piint_(pi/6)^(pi/2)cscucotu du`

`=1/pi(-cscu)]_(pi/6)^(pi/2)=1/pi(-csc(pi/2)+csc(pi/6))`

`-1/pi(-1+2)=1/pi`