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