<s>[INST] What is wrong with this approach evaluating $\int \sec \theta \ d \theta$? When integrating a function like $\sin^m \theta \cdot \cos^n \theta$ where where $m,n$ are nonnegative integers and $n$ is odd, a common approach is to peel off one power of $\cos \theta$ and then rewrite the resulting even power of $\cos \theta$ in terms of $\sin \theta$ using $\cos^2 \theta = 1 - \sin^2 \theta$. What remains is a polynomial in $\sin \theta$, multiplied by $\cos \theta$, so the substitution $u = \sin  \theta$ can be used. Since $\sec \theta =(\sin \theta)^0 (\cos \theta)^{-1}$ and $-1$ is odd, it seems like a good idea to try out the same approach here. 
\begin{align*}
\int \sec \theta \ d \theta 
&= \int \frac{\cos \theta}{\cos^2 \theta} \ d \theta && \text{Splitting off a power of $\cos \theta$}\\
&=\int \frac{ \cos \theta}{1-\sin^2 \theta} \ d \theta && \text{Eliminating the even power of $\cos \theta$} \\
&=\int \frac{1}{1-u^2}\ du &&  \text{Substituting $u=\sin \theta$}\\
&=\frac{1}{2} \int \left( \frac{1}{1+u} + \frac{1}{1-u} \right) \ du&& \text{Partial fractions}\\
&=\frac{1}{2} \left( \log |1+u| + \log |1-u| \right) && \\
&=\frac{1}{2} \log |1-u^2|  && \\
&=\frac{1}{2} \log| \cos^2 \theta| && \text{Restoring $u=\sin \theta$} \\
&=\frac{1}{2} \log (\cos^2 \theta) && \text{Eliminating redundant absolute signs}\\
&= \log \sqrt{ \cos^2 \theta}&& \\
&= \log|\cos \theta|. \\
\end{align*}
I didn't bother adding the customary constant because the domain of $\sec \theta$ is disconnected and so you would actually need to add a piecewise constant function in order to specify the general antiderivative.
My questions are: 

Question 1: Since $\frac{d}{d \theta} \log|\cos \theta| = - \tan \theta$, where does this calculation go wrong?

And also:

Question 2: Can this approach be salvaged?

 [/INST] In your fifth line,
$$\log |1+u| + \log |1-u|$$
should be
$$\log |1+u| - \log |1-u|\ .$$
This then gives
$$\eqalign{
  \frac12\log\Bigl|\frac{1+u}{1-u}\Bigr|
  &=\frac12\log\Bigl(\frac{1+\sin\theta}{1-\sin\theta}\Bigr)\cr
  &=\frac12\log\Bigl(\frac{1+\sin\theta}{1-\sin\theta}
    \frac{1+\sin\theta}{1+\sin\theta}\Bigr)\cr
  &=\log\frac{1+\sin\theta}{\cos\theta}\cr
  &=\log(\sec\theta+\tan\theta)\ .\cr}$$
</s>