Integrating $\int \sec^2(x) \tan(x) dx$ by trig substitution

Question

I know I am supposed to integrate

$$\int \sec^2(x) \tan(x) dx$$

by substituting $u = \tan(x)$ and get $du = \sec^2(x)$. However, why can't I use $u = \sec(x)$, $du = \tan(x) \sec(x)$?

Answer 1

You can use either substitution.

If you use $u = \tan x$, $du = \sec^2 x \,dx$, then you get

$\displaystyle\int \sec^2 x \tan x \,dx = \int u\,du = \dfrac{1}{2}u^2+C = \dfrac{1}{2}\tan^2 x + C$ for some constant $C$.

If you use $u = \sec x$, $du = \sec x\tan x \,dx$, then you get

$\displaystyle\int \sec^2 x \tan x \,dx = \int u\,du = \dfrac{1}{2}u^2+C' = \dfrac{1}{2}\sec^2 x + C'$ for some constant $C'$.

These two answers are equivalent. Since $\sec^2 x = \tan^2 x + 1$, this just means that $C = C'+ \dfrac{1}{2}$.