Evaluation of $\int \sec^3 (x)dx$ How Can I evaluate $\displaystyle \int \sec^3 (x)dx$
(Without Using Weierstrass Substution or Integration by parts.)
$\bf{My\; Try::}$ 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 $\displaystyle dx = -dt$. So $\displaystyle I = -\int \frac{1}{\sin^3 t}dt = -\int\frac{1}{2\sin^3\left(\frac{t}{2}\right)\cdot \cos^3 \left(\frac{t}{2}\right)}dt$
Now How can I solve after that
Help me
Thanks

Hint:
Substitite
$$\sec (x)= \cosh (u) $$
$$\tan(x) = \sinh(u)$$
$$\sec^2 (x) \, \mathrm dx = \cosh (u) \,\mathrm du$$
$$\mathrm dx = \frac1{\cosh(u)}  \, \mathrm du$$
Edit: Why was this downvoted?
Here is the entire process 
$$\begin{align}
\int \sec^3 x \, \mathrm dx &{}= \int \cosh^2 u\,\mathrm du \\
&= \frac{1}{2}\int ( \cosh 2u +1) \,\mathrm du \\
&= \frac{1}{2} \left( \frac{1}{2}\sinh2u + u\right) + C\\
&= \frac{1}{2} ( \sinh u \cosh u + u ) + C \\
&= \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln|\sec x + \tan x| + C
\end{align}
$$
