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

3 Answers
Oct 16, 2014

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.

Apr 27, 2018

#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#

Apr 27, 2018

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"#