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Using the method of replacing show that ∫tan(x)=dx=In|sec(x)|+constantSee correction

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jordsonsmith | Student, Undergraduate | eNoter

Posted October 28, 2012 at 12:01 AM via web

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Using the method of replacing show that ∫tan(x)=dx=In|sec(x)|+constant

See correction

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juniorsilvamath | Student, Undergraduate | Honors

Posted October 28, 2012 at 12:15 AM (Answer #1)

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∫tan(x)=dx=In|sec(x)|+constante

∫tan x dx = ∫(sin x/cos x)dx
set
u = cos x.
then we find
du = - sin x dx

substitute du = -sin x, u = cos x
dx = - ∫(-1) sin x dx/cos x

= -∫du/u

Solve the integral

= - ln |u| + C

substitute back u=cos x

= - ln |cos x| + C
Q.E.D.

2. Alternate Form of Result

∫tan x dx = - ln |cos x| + C
= ln | (cos x)-1 | + C
= ln |sec x| + C
Therefore:
∫tan x dx = - ln |cos x| + C = ln |sec x| + C

 

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lfryerda | High School Teacher | (Level 2) Educator

Posted October 28, 2012 at 4:56 AM (Answer #2)

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To evaluate this integral, it is best to remember that `tanx=sinx/cosx` , which suggests the substitution `u=cosx` .  This means that `du=-sinxdx` , so the integral becomes:

`int tanxdx`

`=int sinx/cosx dx`

`=-int 1/udu`   now integrate using logarithm rule

`=-ln|u|+C`  where C is a constant of integration

`=-ln|cosx|+C`   now switch negative 1 to exponent

`=ln|cosx|^{-1}+C`   use trig definition of `secx` .

`=ln|secx|+C`

The integral evaluates to `ln|secx|+C` .

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