integration

Observe that ∫ [sec^2(x)] dx = tan x
When integrating [tan^2(x)sec^2(x)] , consider tan^3 (x)
d tan^3 (x)/dx
= 3 tan^2 (x) sec^2 (x)
d [tan^3 (x)/3] / dx = tan^2(x)sec^2(x)
So ∫ [tan^2(x)sec^2(x)] dx = tan^3 (x) / 3 +C

same for the second question, observe that d secx /dx = tan x sec (x)
Then d sec^2 (x) / dx = 2tan x sec^2(x)
d [sec^2(x) /2] /dx = tan x sec^2 (x)
So ∫ [tan(x)sec^2(x)] dx = sec^2(x) /2 +C