Question

evaluate the integral

Answer 1

dibs. lol jk

Answer 2

\[\int\limits_{0}^{\pi/12} (1+e ^{\tan3x})(\sec ^{2}3xdx)\]

Answer 3

let u = tan(3x) --> du = 3sec^2(3x) dx

Answer 4

= \[= 1/3 \int\limits_{0}^{\pi/12} (1 + e^{u}) du\]

Answer 5

integrate for u. then plug u = tan(3x) back in. let me know if i should keep going or if it's enough

Answer 6

actually. there is something wrong with this.

Answer 7

you can keep going

Answer 8

the limits are not 0 to pi/12. they are tan(3(limit)). but when we re-plug u = tan(3x) the limits go back to 0 to pi/12 so we don't really need to worry about it

Answer 9

\[=1/3 ( u + e^{u}) = 1/3 (\tan(3x) + e^{\tan(3x)}) with the \limits 0 \to \pi/12 \]

Answer 10

lol the latex didnt work for me

Answer 11

is it understandable?

Answer 12

im doing the math, all my choices still have e in them but wouldn't e go away

Answer 13

what do you mean by go away?

Answer 14

choices are e/3, e, 3e, or -e/3

Answer 15

when its calculated wouldn't e become a number

Answer 16

answer is 1/3 ( tan(3pi/12) + e^(tan(3pi/12)) - tan(3(0)) - e^0)

Answer 17

sorry about that. i plugged it into the u version not the x version >.<

Answer 18

= 1/3 ( 1 + e - 0 - 1 ) = e/3

Answer 19

thank you

Answer 20

glad i could help :)