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?

A less simple (but not complicated) solution using the tangent half-angle substitution $t=\tan(\frac x2)$. $$I=\int \sec^3x\,dx=\int\frac{2 \left(t^2+1\right)^2}{\left(1-t^2\right)^3}\,dt$$ Now, partial fraction decomposition $$\frac{2 \left(t^2+1\right)^2}{\left(1-t^2\right)^3}=\frac{1}{2 (t+1)}-\frac{1}{2 (t+1)^2}+\frac{1}{(t+1)^3}-\frac{1}{2 (t-1)}-\frac{1}{2
   (t-1)^2}-\frac{1}{(t-1)^3}$$ Integrating and simplifying $$I=\frac{t^3+t}{\left(t^2-1\right)^2}+\frac 12 \log\Big(\frac{1+t}{1-t}\Big)$$
