How do you integrate \(\int \sec^{2n+1} x \, dx\)?

[nhrock3]

ok but i get the same integral s before

 

[hunt_mat]

Like I said, I can't think of any other way of doing it other than the way I have mentioned. Perhaps your lecturer can.

 

[Gib Z]

Integrating I_n = \int \sec^{2n+1} x dx:

1. Integrate by parts to get a recurrence relation.
With u=\sec^{2n-1} x , du = (2n-1) \sec^{2n-2} x \cdot \sec x \tan x dx and dv = \sec^2 x dx, v= \tan x we have
I_n = \int \sec^{2n+1} x dx = \sec^{2n-1} x \tan x - \int (2n-1) \sec^{2n-1}\cdot \tan^2 dx
= \sec^{2n-1} x \tan x - (2n-1) \biggl( \int sec^{2n+1}x dx- \int \sec^{2n-1}x dx \biggr)
= \sec^{2n-1} x \tan x -(2n-1)\left( I_n - I_{n-1} \right)
I_n = \frac{\sec^{2n-1} x \tan x +(2n-1)I_{n-1}}{2n}

This eventually reduces the problem down into I_1 = \int \sec x dx = \int \frac{\sec x (\sec x + \tan x)}{\sec x + \tan x} dx = \log |\sec x + \tan x| + C

2. Make a substitution, then integrate a rational function.
I_n = \int \frac{\cos x}{\cos^{2n+2} x}{dx} = \int \frac{ d(\sin x) }{(1-\sin^2 x)^{n+1} } = \int \frac{1}{(1-u^2)^{n+1}} du
Now it is a simple but tedious exercise in partial fractions.

3. Let x= \sin^{-1}\left(\tanh t\right), then
I_n = \int \cosh^{2n} t dt

\cosh^{2n} x= \left( \frac{e^x + e^{-x} }{2} \right)^{2n}
= \frac{1}{2^{2n}}\left( \binom{n}{0} e^{2nx} + \binom{n}{1} e^{2(n-1)x} + \binom{n}{2} e^{2(n-2)x} + \cdots + \binom{n}{n-2} e^{-2(n-2)x} + \binom{n}{n-1} e^{-2(n-1)x} + \binom{n}{n} e^{-2nx} \right)
= \frac{1}{2^{2n-1}} \left( \binom{n}{0} \cosh (2nx) + \binom{n}{1} \cosh (2(n-1)x) + \binom{n}{2} \cosh ( 2(n-2)x ) + \cdots

So integrating term by term:
I_n = \frac{1}{2^{2n}} \left( \binom{n}{0} \frac{\sinh (2nt)}{n} + \binom{n}{1} \frac{\sinh (2(n-1)t) }{n-1} + \binom{n}{2} \frac{ \sinh (2(n-2)t)}{n-2} + \cdots + C
where t= \tanh^{-1} \left(\sin x\right).