Question

How do you integrate `tanx/(1+cosx)`?

Answer 1

`inttanx/(1+cosx)dx=intsinx/(cosx(1+cosx))dx`

let

`u=cosx`
`du=-sinxdx`

partial fractions

`-int(du)/(u(u+1))=-int(1/u-1/(u+1))du`

`-int(1/u-1/(u+1))du=int(du)/(u+1)-int(du)/u`

`-int(du)/u=-ln|u|`

For `int(du)/(u+1)`

set
`w=u+1`
`dw=du`

`int(dw)/w = ln|w|`

Plug back in for u and x

`ln|w|= ln|u+1|`

So far we have

`ln|u+1|-ln|u|`

Remember `u=cosx`

`ln|cosx+1|-ln|cosx|`

And `+ C`

`inttanx/(1+cosx)dx=ln|cosx+1|-ln|cosx|+C`