What is the antiderivative of #tan(x)#?
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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.
#int# #tanx# #dx=ln|secx|+C#
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#
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"#
