## Question

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

## Answers

The answer is

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

#"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#

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

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#