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# How to determine the integral `int sec x dx`

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How to determine the integral `int sec x dx`

Posted by lxsptter on October 4, 2013 at 4:17 AM via web and tagged with calculus, math

College Teacher

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The integral `int sec x dx` has to be determined.

The integral of sec x can be solved with the use of an indirect method.

Multiply sec x by `(sec x+ tan x)/(sec x + tan x)`

`int sec x dx`

= `int (sec x*(sec x+ tan x))/(sec x + tan x) dx`

= `int (sec^2 x + sec x*tan x)/(sec x + tan x) dx`

Let `sec x + tan x = y`

`dy/dx = sec x*tan x + sec^2x`

=> `dy = sec x*tan x + sec^2x dx`

This gives:

`int (sec^2 x + sec x*tan x)/(sec x + tan x) dx`

= `int 1/ y dy`

= `ln y`

Substitute `y = sec x + tan x`

= `ln(sec x + tan x)`

The integral `int sec x dx = ln|sec x + tan x| + C`

Posted by justaguide on October 4, 2013 at 4:28 AM (Answer #1)

College Teacher

(Level 2) Distinguished Educator

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The integral `int sec x dx` has to be determined

`int sec x dx`

= `int 1/cos x dx`

= `int cos x/cos^2x dx`

= `int cos x/(1 - sin^2x) dx`

Let `sin x = y`

`dy/dx = cos x`

=> `dy = cos x*dx`

`int cos x/(1 - sin^2x) dx`

= `int 1/(1 - y^2) dy`

= `int 1/((1 - y)(1 + y)) dy`

= `int -1/((y - 1)(y + 1)) dy`

= `int (1/2)*(y - 1 - (y + 1))/((y - 1)(y + 1)) dy`

= `(1/2)int 1/(y+1) - 1/(y-1) dy`

= `(1/2)(int 1/(y+1) dy - int 1/(y-1) dy)`

= `(1/2)(ln(y + 1) - ln(y - 1))`

= `(1/2)*ln((y + 1)/(y - 1))`

substitute `y = sin x`

= `(1/2)*ln((sin x + 1)/(sin x - 1))`

The integral `int sec x dx = (1/2)*ln((sin x + 1)/(sin x - 1)) + C`

Posted by justaguide on October 4, 2013 at 4:56 AM (Answer #1)

High School Teacher

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`intsec(x)dx=int(sec(x)(sec(x)+tan(x)))/(sec(x)+tan(x))dx`

LetÂ  t=sec(x)+tan(x)

dt=(sec(x)tan(x)+sec^2(x))dx

Thus

`intsec(x)dx=int((sec^2(x)+sec(x)tan(x))dx)/(sec(x)+tan(x))`

`=int1/tdt`

`=ln|t|+c`

`=ln|sec(x)+tan(x)|+c`

Posted by aruv on October 4, 2013 at 5:42 AM (Answer #2)