Question

How do you find the antiderivative of `int tanx dx`?

Answer 1

`lnabssecx+C" "` or `" "-lnabscosx+C`

Explanation:

Rewrite `tanx` using its form in `sinx` and `cosx`:

`inttanxdx=intsinx/cosxdx`

Now, we can use the substitution `u=cosx`. This implies that `du=-sinxdx`.

`intsinx/cosx=-int(-sinx)/cosxdx=-int(du)/u`

This is a common and valuable integral to recognize:

`-int(du)/u=-lnabs(u)+C`

Back-substituting with `u=cosx`:

`-lnabsu+C=-lnabscosx+C`

Note that this can be rewritten using logarithm rules, with the negative on the outside being brought into the logarithm as a `-1` power.

`-lnabscosx+C=lnabs((cosx)^-1)+C=lnabssecx+C`