Integrating sec, sech, cosec and cosech (cosh)

By Sarah Shales on November 2, 1998 :

Dear Nrich,

Is there a common pattern or theme in the integrations of :

sec x, sech x, cosec x and cosech x .

Thank you

Sarah Shales

By Simon Munday (sjm78) on November 10, 1998 :

Hi Sarah,

Thanks for the question. Yes, there is a pattern, but strangely, you have to use the derivatives of the functions you're trying to integrate. I'll take sec x as an example (the others work in essentially similar ways).

d/dx(sec x) = sec x tan x, and also 1 + tan² x = sec² x.

Now we can write sec x = (sec x tan x) / tan x

=[d/dx(sec x)] / [sec² x - 1]^1/2 .

So Int{sec x dx} = Int{1/[u² - 1]^1/2 du}

where I have made the substitution u = sec x. Now the RHS is a standard integral (it gives some inverse trig or hyperbolic function), which I'm sure you can do (or look up), and you will have the answer. The same thing works for the other three, but be careful of plus and minus signs being different!

I hope that answers your question,

By David (djs61) on November 22, 1998 :

I remember trying to evaluate these integrals when I was at school and generally getting confused, particularly with

cosech x

. I think I also found it difficult to understand why things like

\arctan

and

\ln

kept appearing in the results.

It's possible to appeal to definitions and write the functions in terms of (possibly complex) exponentials. Thus

$\int (sech x) dx = \int (2 / e^{x} + e^{- x}) dx$

and

$\int (cosech x) dx = \int (2 / e^{x} - e^{- x}) dx$

This way all the ratios look much the same, modulo a few minus signs, and the inevitable $i$ 's in the exponent for the circular functions.

The difference of two between the arguments of the exponential terms suggests some sort of attack by obtaining a quadratic term $u^{2} + 1$ or $u^{2} - 1$ in the denominator, and indeed an appropriate substitution ( $u = \exp (x)$ or $u = \exp (- x)$ ) does the business: standard integrals appear, inverse circular (or hyperbolic) functions are invoked, and another pattern emerges.

This works for all these functions and explains the appearance of $\arctan$ or $artanh$ in all of these integrals. (Both of these can also be written as natural logarithms.) However, the $cosech$ integral is a little fragile (since $\sinh 0 = 0$ ) and, if you evaluate it this way, you must be careful which substitution you pick: you need $u = e^{x}$ for $x < 0$ and $u = e^{- x}$ for $x > 0$ . (You might like to think about why this is necessary.)

David