Question

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

Answer 1

`tan^2(x/2)+C_1" "`or `sec^2(x/2)+C_2`

Explanation:

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`

Answer 2

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.