I already know that there are multiple answers of integrating $\int_{}\sec^3x$. My textbook uses an integration by parts approach and I also have seen a partial functions approach as well.

I stumbled upon this post and one of the users @N.S. posted an interesting approach that used substitution, but I wasn't able to work the problem out completely...Here is where I got to

$$\int_{} \sec^3(x) = \int \frac{1}{\cos^3x} * \frac{\cos(x)}{\cos(x)} = \int \frac{\cos(x)}{\cos^4(x)} = \int \frac{\cos(x)}{(1- \sin^2(x))^2}dx $$

Apply substitution with $u = \sin(x), du = \cos(x)$:

$$\int \frac{du}{(1-u^2)^2} = \int \frac{du}{1 - 2u^2 + u^4} = \frac{1}{\frac{2}{3}u^3 - \frac{u^5}{5}} $$

But this just seems wrong and nowhere close to the answer of $\frac{1}{2} (\sec(x) \tan(x) + ln| \sec(x) + \tan (x)|) + C$

  • $\begingroup$ Your last integration is not correct. $\endgroup$ – Severin Schraven Feb 28 '19 at 20:54
  • $\begingroup$ The answer in the post did not show the full work, but just claimed that the theory of trig integrals says that if cos(x) appears as an odd power, then the substitution $u = \sin(x)$ should work....I was just trying to see if that was indeed true $\endgroup$ – Evan Kim Feb 28 '19 at 21:00
  • $\begingroup$ From the integral $\int (1 - u^2)^{-2} du$ it's natural to decompose using partial fractions. You can save some effort by using the fact that the integrand is even. $\endgroup$ – Travis Willse Feb 28 '19 at 21:03

Keep one cosine, and convert the others to sine. Then substutute $u = \sin x, du = \cos x\;dx$. \begin{align} \int \sec^3 x\;dx &= \int\frac{\cos x\;dx}{\cos^4 x} = \int\frac{\cos x\;dx}{\left(1-\sin^2 x\right)^2} \\&= \int \frac{du}{(1-u^2)^2} \end{align} so it is reduced to a rational function. All that was correct. But you did not integrate the rational function correctly. Perhaps use the method of partial fractions? $$ \frac{1}{(1-u)^2(1+u)^2} = \frac{1}{4(1+u)}+\frac{1}{4(1+u)^2} +\frac{1}{4(1-u)}+\frac{1}{4(1-u)^2} $$

  • $\begingroup$ Oh okay, that makes sense. I thought there was an alternate method instead of partial fractions that worked (so was not trying method of partial fractions) $\endgroup$ – Evan Kim Feb 28 '19 at 21:30

How did you perform that final integration? It looks like you've assumed some nonexistent identity. Go with partial fractions (for $\frac{1}{1-u^2}$ first, then square it). You'll find $(\sec x+\tan x)^2=\frac{1+\sin x}{1-\sin x}$ useful.

  • 1
    $\begingroup$ Okay, I will do that $\endgroup$ – Evan Kim Feb 28 '19 at 21:30

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