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?

$\bf{My\; Solution::}$ Let $\displaystyle I = \int \sec^3 x dx = \int \frac{1}{\cos^3 x} dx = \int\frac{1}{\sin^3\left(\frac{\pi}{2}-x\right)}dx$
Now Let $\displaystyle \left(\frac{\pi}{2}-x\right) = t\;,$ Then $dx = -dt$
So Integral $\displaystyle I = -\int\frac{1}{\sin^3 t} dt =-\frac{1}{8}\int \frac{1}{\sin^3 \frac{t}{2}\cdot \cos^3 \frac{t}{2}}dt  = -\frac{1}{8}\int\frac{\sec^6 \frac{t}{2}}{\tan^3 \frac{t}{2}}dt = \frac{1}{8}\int\frac{\sec^4 \frac{t}{2}\cdot \sec^2 \frac{t}{2}}{\tan^3 \frac{t}{2}}dt$
Now Let $\displaystyle \tan \frac{t}{2} = u\;,$ Then $\displaystyle \sec^2 \frac{t}{2}dt = 2du$
So Integral $\displaystyle I = -\frac{1}{4}\int\frac{(1+u^2)^2}{u^3}du = -\frac{1}{4}\int\frac{1+u^4+2u^2}{u^3}du = -\frac{1}{4}\int \left[u^{-3}+u+2u^{-1}\right]du$
