help with the integral of tan

Question

Okay, so my but shows e^ integral tanx to be e^ln|secx|. 

 

as far as I know, the integral of tanx is -ln|COSx|. 

 

how do they do that?

Answer 1

∫tan(θ)dθ = ∫(sin(θ)/cos(θ))dθ 

 

use substitution, u = cos(θ); du = -sin(θ)dθ

 

∫(sin(θ)/cos(θ))dθ = -∫(1/u)du

 

= -ln |u| + C

 

substituting u = cos(θ)

 

= -ln |cos(θ)| + C

 

 

-ln |cos(θ)| = ln |sec(θ)| due to the logarithm rule ln |1/x| = - ln |x|.

 

We can confirm this result by finding the derivation. We should come up with the original function.

 

d(-ln |cos(θ)| + C)/dθ = d(-ln |cos(θ)|)/dθ + d(C)/dθ

 

= d(-ln |cos(θ)|)/dθ 

 

using the chain rule,

 

dz/dθ = dz/dy*dy/dθ, y = cos(θ)

 

d(-ln |cos(θ)|)/dθ = d(-ln |y|)/dy * d(cos(θ))/dθ = -1/y * -sin(θ) 

 

= -1/cos(θ) * -sin(θ)  = tan(θ)

Answer 2

Yes. You are correct about the formula for the integral being -ln|cos x|+C, since it is a routine  u-substitution with u = cos x and du = -sin x dx

 

∫ tan x dx

 

= ∫ (sin x)/(cos x) dx

 

= ∫ -du/u

 

= -ln|u| + C

 

= -ln|cos x| + C

 

Because sec x = 1/cos x, we get

 

-ln|cos x| + C = ln|sec x| + C

Answer 3

Now - ln|cos(x)| = ln|cos(x)^-1|

                        = ln|1/cos(x)|

                        = ln|sec{x}|

Answer 4

So -ln|cosx| is equal to ln|secx| due to secx = 1/cosx.

 

See if you put ln|secx| = ln|1/cosx|, then you can use the rule of logs where a division becomes subtraction:

 

ln|1/cosx| = ln|1| - ln|cosx| which in turn gives you -ln|cosx| as requried. (Because ln|1| = 0)