What is #int (tanx)/sqrt(sec^2x-4)dx#?

I got:

#int (tanx)/(sqrt(sec^2x - 4))dx = 1/2 "arc"sec(secx/2) + C#

Try multiplying by #(secx)/(secx)#.

#= int (secxtanx)/(secxsqrt(sec^2x - 4))dx#

Then, let #u = secx# so that #du = secxtanxdx#. We get:

#= int 1/(usqrt(u^2 - 4))du#

Now this looks like a form where trig substitution would work. Let #u = 2secs# so that #du = 2secstansds#. Then:

#= int 1/(2secs*2tans)(2secstansds)#

#= 1/2 int ds = s/2#

#= 1/2 "arc"sec(u/2)#

Now, we back-substitute... Since #u = secx#, we have

#= 1/2 "arc"sec(secx/2)#

Sadly, for simplifications, we cannot undo the #"arc"sec# as #"arc"sec# is not the inverse of #sec/2#. So, we'll have to leave it like this.

#color(blue)(int (tanx)/(sqrt(sec^2x - 4))dx = 1/2 "arc"sec(secx/2) + C)#

If we take the derivative, we should get the original integrand back.

#d/(dx)[1/2 "arc"sec(secx/2)] = 1/2 1/(|secx/2| sqrt((secx/2)^2 - 1)) * 1/2 secxtanx#

#= 1/4 (2secxtanx)/(|secx|sqrt(sec^2x/4 - 1))#

For #x > 0#:

#= (tanx)/(2sqrt(sec^2x/4 - 1))#

#= (tanx)/(sqrt(4 (sec^2x/4 - 1)))#

#= (tanx)/(sqrt(sec^2x - 4))# #color(blue)(sqrt"")#