The Student Room Group
Reply 1
secxdx=lnsecx+tanx+C\int \sec x \, \mathrm{d}x = \ln | \sec x + \tan x | + C
Reply 2
FWoodhouse
secxdx=lnsecx+tanx+C\int \sec x \, \mathrm{d}x = \ln | \sec x + \tan x | + C


Francis is always in the zone :biggrin:
Reply 3
Lusus Naturae
Francis is always in the zone :biggrin:


I know math-fu.
Reply 4
The above.

If your interested in how to do it it requires a strange method like multiplying by sec squared + tan squared/ sec squared + tan squared or something.

Whatever it is it's not something that would ever occur to you to do if its the one I'm thinking of.
Reply 5
As the result results, multiply the integrand by "sec x + tan x" to get int f'(x)/f(x) dx.

Alternatively, use the substitution "tan (t/2)" (find a formula for sec x not involving trigonometric functions).
Yeah, everyone's already said it, but I'll repeat in nice maths notation. :p:

secxdx=secx(secx+tanx)secx+tanxdx=sec2x+secxtanxsecx+tanxdx\huge \int \sec x \, \text{d} x = \int \frac{\sec x(\sec x + \tan x)}{\sec x + \tan x} \, \text{d} x = \int \frac{\sec ^2 x + \sec x \tan x}{\sec x + \tan x} \, \text{d} x
=ddx(secx+tanx)secx+tanxdx=lnsecx+tanx+c\huge = \int \frac{\frac{\text{d}}{\text{d} x} ( \sec x + \tan x )}{\sec x + \tan x} \, \text{d} x = \ln | \sec x + \tan x | + c