What is the Integral of #tan^2(x)/sec(x) #?

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Answer 1 · 2018-04-03T11:57:26

The answer is #=ln(|tanx+secx|)-sinx+C#

#"Reminder"#

#intsecxdx=ln(tanx+secx)+C#

Therefore, the integral is

#int(tan^2xdx)/(secx)= intcosxtan^2xdx#

#=intcosx(sec^2x-1)dx#

#=int(secx-cosx)dx#

#=ln(|tanx+secx|)-sinx+C#

Answer 2 · 2018-04-03T12:29:39

#int tan^2(x)/(sec(x))dx=lnabs(sec(x)+tan(x))-sin(x)+C#

We have:

#int tan^2(x)/(sec(x))dx#

We can rewrite this as:

#int tan^(2)(x)*(sec(x))^-1dx# Remember that:

#tan^2(x)=sec^2(x)-1#

#=>int (sec^2(x)-1)*(sec(x))^-1dx#

#=>int sec(x)-(sec(x))^-1dx#

#=>int sec(x)dx-int(sec(x))^-1dx#

Here, remember that #intsec(x)dx=lnabs(sec(x)+tan(x))+C#

#=>lnabs(sec(x)+tan(x))-int(sec(x))^-1dx# (You can ignore the #C#.)

Now...

What does #sec^-1(x)# equal to?

First, #(cos(x))^-1=sec(x)#

Therefore, #sec^-1(x)=((cos(x))^-1)^-1=>cos(x)#

Therefore, we now have:

#=>lnabs(sec(x)+tan(x))-intcos(x)dx#

Another thing to remember is that #intcos(x)dx=sin(x)#

Therefore, we now have:

#=>lnabs(sec(x)+tan(x))-sin(x)# Do you #C# why this is incomplete?

#int tan^2(x)/(sec(x))dx=lnabs(sec(x)+tan(x))-sin(x)+C#