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I have a question regarding the following reduction formula:

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

My calculus book states that it is only valid for $n\ge3$. Why is this the case? How does one intuit such a result?

Sure enough, $I_1$ breaks down because of division by zero. But what about $n=2$? And why can't $n=0$ be a base-case? Or even $n=-1$ etc.?

Finally: How do I know for sure it will actually work with all $n\ge3$?

GambitSquared
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  • What does it give for $I_2$? – Angina Seng Feb 14 '18 at 21:08
  • A far better question is to ask why the reduction formula should work for $n = 1$. It doesn't really make sense to expect that it would - the proof of the reduction formula involves $I_{n - 2}$, which looks totally different from $I_n$ if $n \le 2$. So there is absolutely no reason to expect that the same formula holds.... –  Feb 14 '18 at 21:08
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    It's actually not wrong; the formula gives the correct result $\tan x + c$. –  Feb 14 '18 at 21:12
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    There is no factor of $1/2$ when $n=2$, @GambitSquared. – Clayton Feb 14 '18 at 21:12
  • @user296602 I would expect it to work, because the formula seems to involve any $n$... so why does it only work only for some $n$? I don't understand why the fact that the proof uses $n-2$ would make a difference? – GambitSquared Feb 14 '18 at 21:13
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    @GambitSquared: The formula is fine for $n=2$. It falls apart for $n=1$ because of the denominators involving $n$. – Clayton Feb 14 '18 at 21:14
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    Well, the result is trivially false if $n = 0$ and doesn't make any sense when $n = 1$. There is no reason to expect that you can reverse a pattern indefinitely. –  Feb 14 '18 at 21:14
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    Please, there are pencils, and there's paper. If you're too lazy to do the calculations, ask an algebra package, not MSE. –  Feb 14 '18 at 21:16
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    What the book probably meant to say is that the reduction formula is only useful for $n\ge3$. – Barry Cipra Feb 14 '18 at 22:06
  • @user296602 Why false if $n=0$? – BallBoy Feb 14 '18 at 22:41

2 Answers2

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Consider the beginning of the derivation:

$$\int \sec^n(x) dx = \int \sec(x)^{n-2} \sec(x)^2 dx = \sec(x)^{n-2} \tan(x) - \int (n-2) \sec(x)^{n-2} \tan(x)^2 dx.$$

In the case $n=1$, this is true but it is not useful, because you get the same multiple of $I_1$ on the right side as you already had, so you can't isolate $I_1$. You just get the trivial equation $I_1=I_1$.

In the case $n=2$, this technically works (the second term is just zero). But it only works because you've already used the answer in taking the first step (you had to integrate $\sec(x)^2$ to do the integration by parts in the first place). So it doesn't make sense to think of it as part of the recursion for even $n$, instead it is the base case of the recursion for even $n$.

Ian
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  • And what about $I_0$? Why can't $I_0$ be the base case? – GambitSquared Feb 14 '18 at 21:42
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    @GambitSquared Because you need to know $\int \sec(x)^2 dx = \tan(x)+C$ to write down the recursion. Knowing $\int 1 dx = x+C$ does not help you do that. Remember the point is not to derive a neat, clean recursive framework, the point is to calculate trigonometric integrals. – Ian Feb 14 '18 at 21:51
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As the comments are pointing out, it actually does work for $n=2$.

To see why certain values of $n$ are excluded, you have to look at how the formula is proven. Probably the formula is proven by integration by parts. We have $$ \int \sec^n(x)dx = \int\sec^2(x)\sec^{n-2}(x)dx = \tan(x)\sec^{n-2}(x) - \int\tan (x)\frac{d\sec^{n-2}(x)}{dx}dx $$ . The derivative of $\sec^{n-2}(x) = (\sec(x))^{n-2}$ is $(n-2)\sec^{n-3}(x)\sec(x)\tan(x)$ .

Plugging this in gives

$$ \int \sec^n(x)dx = \tan(x)\sec^{n-2}(x) - (n-2)\int\tan^2 (x)\sec^{n-2}(x)dx $$

Now we use the identity $\tan^2(x) = \sec^2(x) - 1$ to make this

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

Adding $(n-2)I_{n}$ to both sides gives

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

Now we get the formula by dividing by $n-1$ -- which we can do for any value of $n$, except $n=1$. (And, of course, I've dropped the $+C$ throughout, but it belongs too.)

So, to answer your question: $n=1$ is special because the division-by-$n-1$ step in the proof won't be possible.

BallBoy
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