Strange behavior in align environment

Question

I am getting this irregular behavior in horizontal alignment of some splitted equations over my document, this is only one example of it. The problem goes in the second split (which has practically the same structure than the first one). I have tried to correct it by putting \\ here and there but with no success. I know that it could have a simple solution, but I can find it. What can I do? Is it a bug of align environment?

\documentclass{article}
\usepackage[utf8x]{inputenc}
\usepackage{amsmath}

\DeclareMathOperator{\di}{d\!}
\DeclareMathOperator{\sen}{\rm sen}

\newcommand{\abs}[1]{\left\lvert\,\strut#1\,\right\rvert} %Absolute value


\begin{document}

\section{Trigonométricas Elementales}
\begin{align}
\int\sen x\,\di x & = -\cos x + C \\
\int\cos x\,\di x & = \sen x + C \\
    \begin{split}
        \int\tan x\,\di x   & = -\ln\abs{\cos x} + C \\
                            & = \ln \abs{\sec x} + C
    \end{split} \\
\int\csc x\,\di x & = -\ln\abs{\csc x- \cot x} + C \\
\int\sec x\,\di x & = -\ln\abs{\sec x+ \tan x} + C \\
    \begin{split}
        \int\cot x\,\di x & = \ln\abs{\cos x} + C \\
                          & = -\ln\abs{\csc x} + C \\
    \end{split} \\
\int\sen^2 x\,\di x & = \frac x2-\frac 14\sen 2x + C \\
\int\cos^2 x\,\di x & = \frac x2+\frac 14\sen 2x + C \\
\int\tan^2 x\,\di x & =\tan x - x + C
\end{align}
\end{document}

Answer 1

"Basically the same" is not the same as "the same"...

\documentclass{article}

\usepackage{amsmath}

\DeclareMathOperator{\di}{d\!}
\newcommand{\sen}{\mathop{\mathrm{sen}}\nolimits}
\newcommand{\abs}[1]{\left\lvert\,\strut#1\,\right\rvert} %Absolute value

\begin{document}

\section{Trigonométricas Elementales}
\begin{align}
  \int\sen x\,\di x & = -\cos x + C \\
  \int\cos x\,\di x & = \sen x + C \\
      \begin{split}
          \int\tan x\,\di x   & = -\ln\abs{\cos x} + C \\
                              & = \ln \abs{\sec x} + C
      \end{split} \\
  \int\csc x\,\di x & = -\ln\abs{\csc x- \cot x} + C \\
  \int\sec x\,\di x & = -\ln\abs{\sec x+ \tan x} + C \\
      \begin{split}
          \int\cot x\,\di x & = \ln\abs{\cos x} + C \\
                            & = -\ln\abs{\csc x} + C %\\ <---
      \end{split} \\
  \int\sen^2 x\,\di x & = \frac x2-\frac 14\sen 2x + C \\
  \int\cos^2 x\,\di x & = \frac x2+\frac 14\sen 2x + C \\
  \int\tan^2 x\,\di x & = \tan x - x + C
\end{align}

\end{document}