integrate tan ^x between 0 and 1?

Answer 1

Try making tan into terms of sin and cos.
tan x dx = (sin x /cos x) dx
Then set:
u = cos x.
du = - sin x dx

= integral(-du/u)

Solve the integral

= - ln |u| + C

substitute back u=cos x

= - ln |cos x| + C

This can also be rewritten use logarithym rules:

= ln |(cos x)^-1| + C
= ln |sec x| + C

To solve the integral from 0 to 1:

ln |sec 0| - ln |sec 1|

ln 1 - ln |sec 1|

0 - ln |sec 1|

This can be rewritten:

-ln |(cos 1)^-1|
ln |cos 1| = .6156

Answer 2

My TI-89 says the integral is

-ln( |cos(x)|), and evaluating from 0 to 1 gives 0.615626.

My TI-83 agrees.

Answer 3

tan x = sin x/cos x
Let H(x) =int tan x dx = int (sin x / cos x) dx = - ln | (cos x )|
H(1) - H(0) = - [ ln cos(1) - ln cos(0)]
So the result = - ln cos(1)

Answer 4

I'll put in the limits later

∫tan(x)dx

-ln|cos(x)| [0, 1]

[-ln|cos(1)| + ln|cos(0)|]

[-0.54 + 1]

0.4597