How do you integrate #int sec^2(x/2)tan(x/2)#?

regonising the function and its derivative is important.

method 1

#d/(dx)tan^2(x/2)=2xx(1/2)tan(x/2)sec^2(x/2)#

so#" "intsec^2(x/2)tan(x/2)dx=tan^2(x/2)+C_1#

method 2

#d/(dx)sec^2(x/2)=(1/2)xx2sec(x/2)xxsec(x/2)xxtan(x/2)#

#=sec^2(x/2)tan(x/2)#

so#" "intsec^2(x/2)tan(x/2)dx=sec^2(x/2)+C_2#

the two solutions can be shown to be consistent by
using the identity

#1+tan^2theta+sec^2theta#

Method 1
If we recognize that #d/dx(tanx) = sec^2x#, then we might try the substitution

#u = tan(x/2)#.

This makes #du = 1/2sec^2(x/2) dx#, and the integral becomes

#2 int u du = u^2 +C = tan^2(x/2) +C#

Method 2

If we recognize that #d/dx(secx) = secx tanx#, then we might try the substitution

#u = sec(x/2)#.

This makes #du = 1/2sec(x/2) tan(x/2) dx#, and the integral becomes

#2 int u du = u^2 +C = sec^2(x/2) +C#

Final Note

A student who has not seem such a result may wonder, "How can both answers be right?"

The answer is in the constant, #C#.

Recall that #tan^2x +1 = sec^2x#

So if we have
#tan^2(x/2) + 5# we can rewrite this as #sec^2(x/2) + 4#.

Our "two" answers have different #C#'s.