How to evaluate $ \int \sec^3x \, \mathrm{d}x$? How to evaluate $ \int \sec^3x \, \mathrm{d}x$ ?
I tried integration by parts and then I got $$\int \sec^3x \, \mathrm{d}x=\sec x \tan x - \int\sec x \tan^2 x \, \mathrm{d}x.$$ Now I'm stuck solving $\int\sec x \tan^2 x \, \mathrm{d}x$? How to solve that and how to solve my initial question? What is the smartest way solving $ \int \sec^3x \, \mathrm{d}x$?
 A: Since
$$\sec^3 x = \frac{1}{\cos^3 x}=\frac{\cos x}{\cos^4 x}=\frac{\cos x}{(1-\sin^2 x)^2},$$
we just need a primitive for
$$ f(t) = \frac{1}{(1-t^2)^2} $$
that is:
$$ F(t) = \frac{1}{4}\left(\frac{2t}{1-t^2}+\log\frac{1+t}{1-t}\right).$$
A: $$
\int\sec^3 x dx = \sec x \tan x - \int \sec x \tan ^2 x dx
$$
using $\tan^2 x+1 = \sec^2 x$ we find
$$
I = \sec x \tan x - \int \sec x \tan ^2 x dx = \sec x \tan x - \int \sec x \left(\sec^2 x-1\right)dx = \sec x \tan x- I + \int \sec x dx
$$
thus we get
$$
2I = \sec x \tan x + \int \sec x dx
$$
now (an interesting derivation and a lot harder than you think to solve try it yourself)
$$
\int \sec x dx = \ln|\sec x + \tan x| + C
$$
thus
$$
I = \frac{1}{2}\left(\sec x \tan x + \ln|\sec x + \tan x| + C\right)
$$
A: Here is a straight way you are looking for:
a) Show that $(\sec x)^{\prime \prime }=2\sec ^{3}(x)-\sec x;$ 
b) Deduce $\int \sec ^{3}x\mathsf{d}x.$
Answer: Recall that $(\sec x)^{\prime }=\sec x\tan x,$ so 
\begin{eqnarray*}
(\sec x)^{\prime \prime } &=&\left( (\sec x)^{\prime }\right) ^{\prime
}=\left( \sec x\tan x\right) ^{\prime }=(\sec x)^{\prime }\tan x+\sec
x\left( \tan x\right) ^{\prime } \\
&=&\sec x\tan x\tan x+\sec x\sec ^{2}x=\sec x(\sec ^{2}x-1)+\sec ^{3}x \\
&=&\sec ^{3}x-\sec x+\sec ^{3}x=2\sec ^{3}(x)-\sec x
\end{eqnarray*}
b) We compute the integral of the second derivative like this
$$
\int (\sec x)^{\prime \prime }\mathsf{d}x=\int (2\sec ^{3}(x)-\sec x)\mathsf{
d}x,
$$
but the integral of a derivative is the function itself! then 
$$
\int (\sec x)^{\prime \prime }\mathsf{d}x=(\sec x)^{\prime }+c=\sec x\tan x+c
$$
therefore
$$
\sec x\tan x+c=2\int \sec ^{3}x\mathsf{d}x-\int \sec x\mathsf{d}x=2\int \sec
^{3}x\mathsf{d}x-\ln (\sec x+\tan x)
$$
from which it follows
$$
\int \sec ^{3}x\mathsf{d}x=\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln (\sec
x+\tan x)+c.
$$
A: This is a well-known problem. Using $ \left( u, v \right) = \left( \sec x, \tan x \right) $ gives $$ \begin {align*} \displaystyle\int \sec^3  \, \mathrm{d}x &= \displaystyle\int \sec x \tan x - \displaystyle\int \sec x \tan^2 x \, \mathrm{d}x \\&= \sec x \tan x - \displaystyle\int \sec^3 x \, \mathrm{d}x + \displaystyle\int \sec x \, \mathrm{d}x. \end {align*} $$If we let our integral be $I$, then, $ 2I = \sec x \tan x + \displaystyle\int \sec x \, \mathrm{d}x = \sec x \tan x + \ln \left| \sec x + \tan x \right| + \mathcal{C}_0 $, so we have $$ \displaystyle\int \sec^3 x \, \mathrm{d}x = \boxed {\dfrac {\sec x \tan x + \ln \left| \sec x + \tan x \right|}{2} + \mathcal{C}}. $$
A: Since I mentioned it in comments, there is a general approach to such problems, the Weierstrass half-angle substitution.
Let $u=\tan \frac{x}{2}$ then $\cos x = \frac{1-u^2}{1+u^2}$ and $dx=\frac{2du}{1+u^2}$. 
So:
$$\int \sec^3 x \,dx = \int \frac{2(1+u^2)^2du}{(1-u^2)^3}$$
You are still going to need to do a partial fraction decomposition.
It's messy. but this trick is worth knowing because it can be used in so many questions.
Cheating to keep this brief, Wolfram Alpha gives the partial fraction decomposition as:
$$\frac{2(1+u^2)^2}{(1-u^2)^3} = \frac{1}{2(u+1)} -\frac{1}{2(u+1)^2} + \frac{1}{(u+1)^3}  - \frac{1}{2(u-1)} - \frac{1}{2(u-1)^2} - \frac{1}{(u-1)^3} $$
