Finding $\int \sec^3x\,dx$ I've tried substitution, replacing $\sec^2x$ with $\tan^2x + 1$, and parts and I just hit dead ends every time... Do you need knowledge of higher-level calculus to solve this?

Notice, $$\int \sec^3x dx$$$$=\int \sec x \sec^2 x dx$$
$$=\int \sqrt{1+\tan^2 x} \sec^2 x dx$$
Let, $\tan x=t\implies \sec^2x dx=dt$
$$=\int \sqrt{1+t^2}dt$$ $$=\frac{1}{2}\left(t\sqrt{1+t^2}+\ln\left|t+\sqrt{1+t^2}\right|\right)+C$$
$$=\frac{1}{2}\left(\tan x\sec x+\ln\left|\tan x +\sec x\right|\right)+C$$
