What happened to the 2!? (integrating trig)

Someone explain please?

The 2 has disappeared in the integration and I am very sad because I don't know why

I know secxtanx integrated = secx which is why it equals sec^2x but the constant 2 is just gone and I wondered if there was an explanation behind this!



Someone explain please?

AS $sec'x=secx\cdot tanx$
your integral form is
$\int f'(x)\cdot f^n(x)\ dx=\frac{f^{n+1}(x)}{n+1}+C$
$2\int secx\cdot sec'x\ dx=2\cdot \frac {sec^2x}{2}+C$

Shouldn't it be $(f(x))^{n+1}$

Because using that notation is like: (for example)
$f(x)=sin(x)$
$f^{2}(x)=sin(sin(x))$

$f^n(x)$ means here $[f(x)]^n$
n can be any real (even 1) except 0 and -1
the nth derivative of f
$f^{(n)}(x)$
or the second derivative
$f^{(2)}(x)=f''(x)$