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: Notice, $$\int \sec^3x dx$$$$=\int \sec x \sec^2 x dx$$
$$=\int \sqrt{1+\tan^2 x} \sec^2 x dx$$
Let, $\tan x=t\implies \sec^2x dx=dt$
$$=\int \sqrt{1+t^2}dt$$ $$=\frac{1}{2}\left(t\sqrt{1+t^2}+\ln\left|t+\sqrt{1+t^2}\right|\right)+C$$
$$=\frac{1}{2}\left(\tan x\sec x+\ln\left|\tan x +\sec x\right|\right)+C$$
A: 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 $$I=\frac{t^3+t}{\left(t^2-1\right)^2}+\frac 12 \log\Big(\frac{1+t}{1-t}\Big)$$
A: $$
\begin{align}
&\int \sec^3x\,dx\\
=&\int \sec x \cdot \sec^2x\,dx\\
\end{align}
$$
Parts: $u = \sec x,\, dv = \sec^2x\,dx$
$$
\begin{align}
&\sec x \tan x - \int \sec x \tan^2x\,dx\\
=\,&\sec x \tan x - \int \sec x\left(\sec^2x-1\right)dx\\
=\,&\sec x \tan x - \int \sec^3x\,dx + \int\sec x\,dx \\
\end{align}
$$
Using $\int\sec x\,dx = \ln\left|\sec x + \tan x\right| + C$, we have:
$$
\begin{align}
\int \sec^3x\,dx &=\sec x \tan x - \int \sec^3x\,dx + \ln\left|\sec x + \tan x\right| + C\\
2\int \sec^3x\,dx &=\sec x \tan x + \ln\left|\sec x + \tan x\right| + C\\
\int \sec^3x\,dx &=\frac{1}{2}\sec x \tan x + \frac{1}{2}\ln\left|\sec x + \tan x\right| + C\\
\end{align}
$$

This is just one way of solving it! This is a notoriously tricky integral. You can also do this with partial fractions, if you know those. I did this solution first because it sticks to more elementary calculus knowledge.
A: Hint: $I = \displaystyle \int \sec xd(\tan x)=\sec x\tan x - \displaystyle \int \tan^2 x\sec xdx= \sec x\tan x - I + \displaystyle \int \sec xdx$
A: Notice that identity $\color{blue}{\sec^2x-1=\tan^2 x}$ and integration by parts give us
\begin{align*}
\int \sec^3 x\,dx-\int\sec x\,dx&=\int\tan^2 x\sec x\,dx=\tan x\sec x-\int\sec x\,dx\\
2\int\sec^3 x dx-\ln|\tan x + \sec x|&=\tan x\sec x+c\\
\color{blue}{\int\sec^3 x\,dx}&\color{blue}{=\frac{1}{2}\tan x\sec x+\frac{1}{2}\ln|\tan x + \sec x|+C}
\end{align*}
Where $C=c/2$, being $c$ and $C$ constants.
A: $\bf{My\; Solution::}$ 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 $\displaystyle I = -\frac{1}{4}\int\frac{(1+u^2)^2}{u^3}du = -\frac{1}{4}\int\frac{1+u^4+2u^2}{u^3}du = -\frac{1}{4}\int \left[u^{-3}+u+2u^{-1}\right]du$
