Integrate $\sec^n(x)\ dx$ for odd n, without using a recursive formula.

Question

$$\int \sec^{2n+1}(x)\ \text{d}x $$ This is not a homework.

I would like someone to suggest a solution for this, without using any recursion formula.

Answer 1

First, substitute with $$ u=\tan{x} $$

$$ I=\int \sec^{2n+1}{x}\,dx=\int (u^{2}+1)^{n-\frac{1}{2}}\,du, n \in \mathbb{N} $$

Then perform a second substitution $$ u=\sinh{\theta }$$

$$ \int (u^{2}+1)^{n-\frac{1}{2}}\,du = \int \cosh^{2n}{\theta }\, d\theta $$

Now here's the big step. One expands the $\cosh{\theta}$ function, either by converting to exponential form, or by expanding the $\cos{\theta}$ function and then using Osborne's rule. I will explain this point later on in the post.

$$ \cosh^{2n}\theta = \frac{1}{2^{2n}}\binom{2n}{n} +\frac{1}{2^{2n-1}}\sum_{k=1}^{n}\ \binom{2n}{n-k}\cosh2k\theta $$

Now we can just integrate term by term.

$$ I=\int \frac{1}{2^{2n}}\binom{2n}{n} +\frac{1}{2^{2n-1}}\sum_{k=1}^{n}\ \binom{2n}{n-k}\cosh2k\theta d\theta $$

$$ = \frac{1}{2^{2n}}\binom{2n}{n}\theta +\frac{1}{2^{2n}}\sum_{k=1}^{n}\binom{2n}{n-k}\frac{\sinh2k\theta }{k}+C $$

Now we have the integral in terms of theta but need to convert back to x. Recall:

$$ u=\tan x=\sinh\theta = \frac{e^{\theta}-e^{-\theta}}{2}$$

$$ \theta = \ln|\tan x + \sqrt{\tan^{2}x+1}| = \ln|\tan x + \sec x| $$

To convert the sinh, note the following:

$$ \sinh(\ln\alpha ) = \frac{\alpha - \alpha^{-1} }{2} $$

Finally, we can write down the answer:

$$ I = \frac{1}{2^{2n}}\binom{2n}{n}\ln|\tan x + \sec x| +\frac{1}{2^{2n+1}}\sum_{k=1}^{n}\binom{2n}{n-k}\frac{(\tan x + \sec x)^{2k}+(\tan x + \sec x)^{-2k}}{k}+C $$

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I've checked this solution and it works for n=0,1 and 2. Say, if you try to differentiate the closed form for n=1 to get back $$\sec^3(x)$$ with Wolfram Alpha, it will give you a big trigonometric function which on simplification does yield $\sec^3(x)$. Alternatively you could try a few limits to check that it works.

It's quite sick really, that one could give any n, say n=15, corresponding to the 31st power of $\sec\theta$, and then one could simply write down its indefinite integral in less than 5 or so minutes.

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I still need to explain the $\cosh\theta$ expansion.

Define the following:

$$ z^{n}=\cos n\theta+i\sin n\theta $$ $$ z^{-n}=\cos n\theta-i\sin n\theta $$

Then,

$$ z^{n}+z^{-n} = 2\cos n\theta $$

Now consider the following:

$$ (z+z^{-1})^{2n} = 2^{2n}cos^{2n}\theta = \binom{2n}{n}+\sum_{k=1}^{n}\ \binom{2n}{n-k}(z^{2k}+z^{-2k})= \binom{2n}{n}+\sum_{k=1}^{n}\ \binom{2n}{n-k}2\cosh 2k\theta$$

Therefore: $$ \cos^{2n}\theta = \frac{1}{2^{2n}}\binom{2n}{n} +\frac{1}{2^{2n-1}}\sum_{k=1}^{n}\ \binom{2n}{n-k}\cos2k\theta $$

Now apply Osborne's rule (rule for converting trig to hyperbolic functions and vice versa - if there is a product of sines, add a negative to convert). Since there are no products of sines, this formula works exactly the same for hyperbolic $\cos\theta$. Thus we arrive at what we wanted:

$$ \cosh^{2n}\theta = \frac{1}{2^{2n}}\binom{2n}{n} +\frac{1}{2^{2n-1}}\sum_{k=1}^{n}\ \binom{2n}{n-k}\cosh 2k\theta $$