Integrating 1/cosx

Hi, I was wondering how you would integrate 1/cosx ??

I thought of re-writing:

(cosx/cos^2x)dx = (cosx/[1-sin^2x])dx
Now how would this be integrated? It looks like integration by parts would do, but how?

Reply 1

Chewwy

t substitution will do the job. alternatively write it as you have done, then let u = sinx

Reply 2

obaydi

rewrite as secx(secx+tanx)/(secx+tanx) and use substitution u = secx+tanx

Reply 3

coldplasma

1 / cosx = sec x

If you integrate sec x, you get ln | secx + tanx |.

Reply 4

tinypony

1 / cosx = sec x

If you integrate sec x, you get ln | secx + tanx |.

That is easy to say, but how would you do so? Mathematica gives a result which is found by using the half-angle formulae.

Reply 5

gyrase

chrisjorg

That is easy to say, but how would you do so? Mathematica gives a result which is found by using the half-angle formulae.

secx
= secx(secx+tanx)/(secx+tanx)
= (sec^2x+secxtanx)/(secx+tanx)

INT secx.dx
= INT (sec^2x+secxtanx)/(secx+tanx).dx

differentiate tanx gives sec^2(x), differentiate secx gives secxtanx, by the rule 'that if the differentials are on the top bunk of the bed whilst the functions are lying cosy on the bottom bunk' we have;

= ln|secx+tanx| + c