Question

i am having problems with integrals, the problem is (sec(x))^2 between pi/6 and -pi/6. i know that the integral tan(X) but when i use the first fundamental theorem of calculus i i get lost, i get the last step and i get lost help please

Answer 1

so, you have \(\Large [\tan x]^{\pi/6}_{-\pi/6} = [\tan (\pi/6) - (-\tan (\pi/6))] = ...\)

Answer 2

yes i have that, its the actual solving that i am having problems with

Answer 3

you know whats tan (pi/6) = ... ?

Answer 4

and i got that because \(\tan (-x) = - \tan x\)

Answer 5

i do believe its .707?

Answer 6

yeah i saw that

Answer 7

tan (pi/6) = sin (pi/6) / cos (pi/6) = .... ?

Answer 8

http://www.mathsisfun.com/geometry/unit-circle.html

Answer 9

use the values from the unit circle

Answer 10

its .6 which is rounded

Answer 11

yes, or you can say its
\(\Large \dfrac{1}{\sqrt 3}\)

Answer 12

so, \(\Large [\tan x]^{\pi/6}_{-\pi/6} = [\tan (\pi/6) - (-\tan (\pi/6))] = ... \\ \Large 2 \tan (\pi/6) = \dfrac{2}{\sqrt 3}\)

Answer 13

thank you, that helped, and very nice profile picture

Answer 14

welcome ^_^
happy to help :)
and since you're new here,
\(\Huge \mathcal{\text{Welcome To OpenStudy}\ddot\smile} \)

Answer 15

thanks lol