2

Recently, a friend of mine asked me for the indefinite integral $$F(x)=\int \sec^3(x)$$ Integrating by parts, letting $u=\sec(x), dv=\sec^2(x)$, we have $$F(x)=\sec(x)\tan(x)-\int \sec(x)\tan^2(x)$$ But now: $$G(x)=\int \sec(x)\tan^2(x)=\int \sec(x)(\sec^2(x)-1)=F(x)-\int\sec(x)$$ And finally(and somewhat magically) $\int\sec(x)=\ln(|\sec(x)+\tan(x)|)$, which gives us: $$F(x)=\sec(x)\tan(x)-(F(x)-\ln(|\sec(x)+\tan(x)|)))$$ $$\implies F(x)=\frac{\sec(x)\tan(x)+\ln(|\sec(x)+\tan(x)|)}{2}$$

The last integral was written as a derivation excercise some weeks ago, so there is where my motivation comes from. But, is there a logical way to come up with it?

chubakueno
  • 5,683
  • 1
  • 23
  • 43
  • 1
    http://math.stackexchange.com/questions/154900/indefinite-integral-of-secant-cubed –  May 26 '14 at 06:15
  • @T.Bongers Thank you for the link. I was specifically asking for the thought process of $\int \sec(x)$, but I also found it in the related links of that post. Partial fractions was a method I dind't knew for integration! Do I close it? – chubakueno May 26 '14 at 06:21
  • 1
    http://en.wikipedia.org/wiki/Integral_of_secant_cubed – lab bhattacharjee May 26 '14 at 06:25
  • 3
    Also be sure to check out the solution using the magical Weierstrass substitution! – David H May 26 '14 at 06:32
  • 1
    This integral is also easy to remember, since its simply the arithmetic mean of the derivative of $\sec(x)$ and integral of $\sec(x)$... – stochasm May 26 '14 at 06:33
  • @DavidH That promises to be the ultimate tool for trigonometric integration :D – chubakueno May 26 '14 at 06:42
  • @chubakueno It's an excellent ace to have up your sleeve. It'll also give you an edge over other students since calculus's classes sadly almost never mention it. – David H May 26 '14 at 07:00
  • for integral of $\sec x$ just let $u = \sec x + \tan x$. I think this is easiest, however, I just pushed the magic back one level. So, is this really better than just knowing the antiderivative of $\sec x$ from memory? Maybe a little... – James S. Cook May 26 '14 at 07:29

0 Answers0