TITLE: Mistake in alternative derivation of $\int \sec(x) \ dx$
QUESTION [5 upvotes]: For practice, I decided to calculate the integral of $\sec(x)$ in my own way. I did it like this:
$$\int \sec(x) \ dx$$
$$= \int \frac{dx}{\cos(x)}$$
$$= \int \frac{1}{y} \frac{dx}{dy} \ dy \ \ \ \ [y=\cos(x)]$$
$$= \int \frac{1}{y \frac{dy}{dx}} \ dy$$
$$= \int \frac{1}{y \cdot y^{\prime}} \ dy$$
$$= \int \frac{y^2 + {(y^{\prime})}^{2}}{y \cdot y^{\prime}} \ dy \ \ \ \ [1 = \cos^{2}(x)+(-\sin(x))^2]$$
$$= \int \frac{y}{y^{\prime}} \ dy \ + \int \frac{y^{\prime}}{y} \ dy$$
$$= \int y \frac{dx}{dy} \ dy \ + \ln(|y|)$$
$$= \int \cos(x) \ dx \ + \ln(\cos(x))$$
$$= \sin(x) + \ln(|\cos(x)|)+c$$
I understand that this is incorrect. But I don't understand where I made the mistake. Why is this procedure wrong?

REPLY [7 votes]: You have
$$\int\frac{y'}{y}dy$$
which you integrate as $\ln y$. This is
$$\int\frac{dy/dx}{y}dy.$$
How do you get this as $\ln y$? I can see that
$$\int\frac{y'}{y}dx$$
does integrate to $\ln y$, but that is not what you have.