Can any boards nerd PLease walk me step by step through integration please !!

fatbastard

ok .. I can do normal integration. It's just those damn fraction integration

ok so

int { cos(x) /tan (x) } for example ... cheers !!

ray giraffe

use tan(x)= sin(x)/cos(x)

MathsArthur

I tried using u=sec x then du = sec x tan x dx and you get

INT cos x dx/tan x = INT dx/sec x tan x = INT du/sec^2 x tan^2 x

= INT du/sec^2 x (sec^2 x -1) = INT du/u^2(u^2-1)

= INT ( -1/u^2 + 1/(u^2-1))du

= INT ( -1/u^2 + 1/(u+1)(u-1) )du

= INT ( -1/u^2 + 1/2(u-1) - 1/2(u+1) )du

= 1/u + ln(u-1)^(1/2) - ln(u+1)^(1/2)

= cos x + ln((sec x - 1)^(1/2) / (sec x + 1)^(1/2))

but this doesn't work when you re-differentiate it to get cos x/tan x

any clues as to what the problem is?

MathsArthur

sorry. i just checked the differentiation and it does work.

for y = cos x + ln ((sec x - 1)^(1/2)/(sec x + 1)^(1/2))

dy/dx = -sin x + (sec x tan x)/2 * ( 1/(sec x-1) - 1/(sec x+1) )

= -sin x + (sec x tan x) * ( 1/((sec x - 1)(sec x + 1)) )

= -sin x + (sec x tan x) * ( 1/(sec^2 x - 1) )

= -sin x + (sec x tan x) * ( 1/tan^2 x)

= -sin x + sec x/tan x

= (-sin^2 x + 1)/sin x

= cos^2 x/sin x = cos x/tan x

as required. anyone know a simpler way to do this.