5
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ This is quite an interesting approach. I can't immediately see what's wrong but I'll keep looking. $\endgroup$ – Jam Aug 23 '18 at 16:17
  • 1
    $\begingroup$ The mixture of $y, y'$ and $dy$ makes it a risky approach. $\endgroup$ – Yves Daoust Aug 23 '18 at 16:30
  • $\begingroup$ I get $\int\frac{y'}{y}\mathrm{d}y=\int\frac{-\sin(x)\left(-\sin(x)\right)}{\cos(x)}\mathrm{d}x=-\sin(x)-\ln\left(\cos(\tfrac{x}{2})-\sin(\tfrac{x}{2})\right)+\ln\left(\cos(\tfrac{x}{2})+\sin(\tfrac{x}{2})\right)+C$ so I think splitting the integral in two might make $\int\frac{y}{y'}$ a bit easier but not $\int\frac{y'}{y}$. $\endgroup$ – Jam Aug 23 '18 at 16:38
6
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.