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 $$ u=tanx $$

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

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 function, either by converting to exponential form, or by expanding the cos 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:

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

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

To convert the sinh, note the following:

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

Finally, we can write down the answer:

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

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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, 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 expansion.

Define the following:

$$ z^{n}=cosn\theta+isinn\theta $$ $$ z^{-n}=cosn\theta-isinn\theta $$

Then,

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

Now consider the following:

$$ (z+z^{-1})^{2n} = 2^{2n}cos^{2n}\theta = \binom{2n}{n}+\sum_{k=1}^{n}\ \binom{2n}{n-k}(z^{2k}+z^{-2k})= \binom{2n}{n}+\sum_{k=1}^{n}\ \binom{2n}{n-k}2cosh2k\theta$$

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. Thus we arrive at what we wanted:

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