Question

Show that `int \ csc x \ dx = ln|tan(x/2)| + C`?

Answer 1

We have:

`int \ csc x \ dx = int \ cscx * (cscx-cotx)/(cscx-cotx) \ dx`
`" "= int \ (csc^2x-cscxcotx)/(cscx-cotx) \ dx`

We now use the following substitution:

`u=cscx-cotx => (du)/dx=-cscxcotx+csc^2x`,

And so our integral becomes:

`int \ csc x \ dx = int \ 1/u \ du`
`" "= ln|u| + C`
`" "= ln|cscx-cotx| + C`

This is the calculus part of the question complete, It now remains to show that this solution is equivalent to the given solution;

We have;

`cscx-cotx = 1/sinx-cosx/sinx`
`" "= (1-cos(2*1/2x))/sin(2*1/2x)`
`" "= (1-(1 - 2sin^2(x/2)))/(2sin(x/2)cos(x/2))`
`" "= (2sin^2(x/2))/(2sin(x/2)cos(x/2))`
`" "= (sin(x/2))/(cos(x/2))`
`" "= tan(x/2)`

And so;

`int \ csc x \ dx = ln|cscx-cotx| + C`
`" "= ln|tan(x/2)| + C " "`, QED