TITLE: Integrate ${\sec 4x}$
QUESTION [8 upvotes]: How do I go about doing this? I try doing it by parts, but it seems to work out wrong:
$\eqalign{
  & \int {\sec 4xdx}   \cr 
  & u = \sec 4x  \cr 
  & {{du} \over {dx}} = 4\sec 4x\tan 4x  \cr 
  & {{dv} \over {dx}} = 1  \cr 
  & v = x  \cr 
  & \int {\sec 4xdx}  = x\sec 4x - \int {4x\sec 4x\tan 4xdx}   \cr 
  & \int {4x\sec 4x\tan 4xdx} :  \cr 
  & u = 4x  \cr 
  & {{du} \over {dx}} = 4  \cr 
  & {{dv} \over {dx}} = \sec 4x\tan 4x  \cr 
  & v = {1 \over 4}\sec x  \cr 
  & \int {4x\sec 4x\tan 4xdx}  = x\sec 4x - \int {\sec 4x} dx  \cr 
  & \int {\sec 4xdx}  = x\sec 4x - \left( {x\sec 4x - \int {\sec 4xdx} } \right) \cr} $

I don't know where to go from here, everything looks like it equals 0, where have I went wrong?
Thank you!
EDIT: Is there an easier way to do this?

REPLY [3 votes]: The standard trick is $$\sec(4x)=\frac{(\sec 4x)(\sec 4x +\tan 4x )}{\sec 4x + \tan 4x}=\frac{\sec^2 4x+\sec 4x \tan 4x}{\sec 4x + \tan 4x}$$
Now take $u=\sec 4x + \tan 4x$, we have $du=4 \sec 4x \tan 4x + 4 \sec^2 4x$, so $\sec 4x dx=\frac{du}{4u}$.