Question

What is the antiderivative of `tan(x)`?

Answer 1

Recall:

`int{g'(x)}/{g(x)}dx=ln|g(x)|+C`

(You can verify this by substitution `u=g(x)`.)

Now, let us look at the posted antiderivative.

By the trig identity `tan x={sin x}/{cos x}`,

`int tan x dx=int{sin x}/{cos x}dx`

by rewriting it a bit further to fit the form above,

`=-int{-sin x}/{cos x}dx`

by the formula above,

`=-ln|cos x|+C`

or by `rln x=lnx^r`,

`=ln|cos x|^{-1}+C=ln|sec x|+C`

I hope that this was helpful.

Answer 2

`int` `tanx` `dx=ln|secx|+C`

Explanation:

By `tanx=sinx/cosx`,

`int` `\tan x` `dx=\int` `\frac{sinx}{cos x}` `dx`

Let `u=cosx`. => `\frac{du}{dx}=-sinx` => `dx={du}/{-sin x}`

By substitution,

`=\int` `{sin x}/u cdot{du}/{-sin x}`

By cancelling `sinx`'s,

`=-\int` `{1}/{u}` `du`

By finding an antiderivative,

`=-ln|u|+C`

By plugging `cosx` back in for `u`,

`=-ln|cos x|+C`

By the log property `rln x=ln x^r`,

`=ln|cos x|^{-1}+C`

By `(cosx)^{-1}=1/{cosx}=secx`,

`=ln|secx|+C`

Answer 3

A different approach...

Explanation:

We want to find `inttanxdx`.

`inttanxdx=int(tanxsecx)/secxdx`

Now let `u=secx` and `du=secxtanxdx`. Then

`int(tanxsecx)/secxdx=int1/udu`

This is a standard integral which evaluates to

`lnabsu+"c"=lnabssecx+"c"`