Studying Mathematics Thread

[polymath257]

Much easier is to realize that the derivative of

sec x + tan x

is

sec x tan x + sec^2 x = sec x(sec x + tan x).

So, write

int sec x dx = int [ sec x (sec x + tan x) ]/(sec x + tan x) dx

and do a substitution u= sec x + tan x. The result falls out.

Yeah, that seems to be the better way of arriving at the expected answer, but requires thinking about differentiating secx+tanx in the first place. There's a reason I call solving these problems a form of art. Lots of creative thinking needed. No way I can do all this on my own without any assistance (at this stage, at least).

Here's my solution to the integral of sqrt(1+x^2). The triangle method comes in handy here.

https://pasteboard.co/HiNzLeB.png

I think a lot of these are discovered by simply playing around enough. So, for example, take a derivative of

ln( tan (x/2) )

to get

1/tan(x/2) * sec^2 (x/2) * 1/2

which is

cos(x/2)/sin(x/2) * 1/cos^2 (x/2) * 1/2

which simplifies to

1/[ 2 sin(x/2) cos(x/2) ]

Which, by the double angle formula, is

1/sin(x) = csc(x).

So, the integral of csc(x) is ln ( tan(x/2) ) +C.

But now,

sec(x) = csc (pi/2 - x), so we can do a substitution to find that the integral of sec x is

- ln( tan ( pi/4 - x/2 ) ) +C

A challenge: show this is the same as the 'usual' anti-derivative for sec x.