1
$\begingroup$

Is it correct to say that this reduction formula is valid for all $n$ except $n=1$?

$$I_n=\int\sec^n(x)dx=\frac{1}{n-1}(\sec^{n-2}x\tan x)+\frac{n-2}{n-1}I_{n-2}+C$$

Which means that it would even work for $n\le0$?

Based on the proof using Integration by Parts I can find no reason to believe it is not valid for any value except $n=1$ (because of division by zero). But since I am just a beginner in math I am not sure. Help would be appreciated!

As an example; filling out the reduction formula for $n=0$ we get:

$$I_0=-\sin x\cos x+2I_{-2}$$ $$I_0=-\sin x\cos x+2\int \sec^{-2}{x}dx$$ $$I_0=-\sin x\cos x+2\int\cos^2xdx$$ $$I_0=-\sin x\cos x+x+\frac{1}{2}\sin{2x}$$ $$I_0=x$$

Which of course corresponds with:

$$I_0=\int\sec^0(x)dx=\int{1}dx=x$$

This question is closely related to this question: Why is the reduction-formula for $\int\sec^n(x)dx$ only valid for $n\ge3$?

Bonus-question: what is the deeper reason it breaks down at $n=1$?

$\endgroup$
  • $\begingroup$ $I_1 = \ln |\sec x + \tan x| + C$ $\endgroup$ – Doug M Feb 15 '18 at 0:41
  • $\begingroup$ @DougM Yes, but that doesn't follow from the Reduction Formula, does it? $\endgroup$ – GambitSquared Feb 15 '18 at 0:43
  • 1
    $\begingroup$ To help us figure out what you mean by "deeper reason", what do you think is the "deeper reason" that $\int x^n\;dx = \frac{1}{n+1}x^{n+1}$ breaks down for $n=-1$? $\endgroup$ – bames Feb 15 '18 at 1:03
  • $\begingroup$ @bames Hmmm, good question, I don't really know the answer to that one either, but I agree the principle is probably the same. $\endgroup$ – GambitSquared Feb 15 '18 at 13:24
  • $\begingroup$ If you frame it the right way, even the power rule does not really break down: $\int_1^x y^a dy$ is a continuous function of $(x,a)$ in $(0,\infty) \times \mathbb{R}$. The log rule then comes from sending $b$ to zero in $(x^b-1)/b$. This is one of many reasons to treat definite integrals with variable limits instead of antiderivatives. $\endgroup$ – Ian Feb 15 '18 at 13:34

Your Answer

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

Browse other questions tagged or ask your own question.