Question

help please number 35

http://prntscr.com/bjedaq

Answer 1

@jtug6

Answer 2

show your work please

Answer 3

^ that and 25/27 right? =)

Answer 4

i did those already lol im just stuck on that one D:

Answer 5

wait which one specifically?

Answer 6

35

Answer 7

this might be helpful :

\[\cot^2x = \dfrac{\cos^2x}{\sin^2x} = \dfrac{1-\sin^2x}{\sin^2x} = \csc^2x-1\]

Answer 8

alright 1 sec

Answer 9

oh its one of those identities

Answer 10

You don't need to memorize a lot of identities. Just remember one identity :
\[\sin^2x + \cos^2x=1\]

All other identites follow from this

Answer 11

yeah what ganshie said lol... sorry i didnt read that you alrdy labeled which problem @ the top ;3

Answer 12

Do you know what to do now that you have csc^2(x) - 1?

Answer 13

\[\int\limits_{\pi/6}^{\pi/2} -x-\cot(x)\]
Can you solve this?

Answer 14

\[\cot^2(x)=-x-\cot(x)\]

Answer 15

hmmm wait i think i got it

Answer 16

what'd you get? :D

Answer 17

rad 3 - pi/3

Answer 18

hmm. That isn't what I got. Show your work like ganshie said first ;p

Answer 19

oh, wait. im sorry xD i used pi/4 not pi/2 im not wearing my contacts, sorry! 1 sec lemme re evaluate haha

Answer 20

lool tyt

Answer 21

yep! that's what I got too. but still i'd like to see your work. what did you end up doing?

Answer 22

okie one sec

Answer 23

mhmmmmm! great! that works. although you evaluated it correctly you wrote pi/2 to pi/2 i think

Answer 24

pi /6 to pi /2 sorry my writing isnt that neat lol Dx

Answer 25

some professors are stingy and may or may not mark that off. ;p

Answer 26

ah its fine just be careful on an actual exam

Answer 27

yesh D: my exam is this thursday

Answer 28

good luck! i got a calc 3 exam on partial derivatives/1st-2nd deriv tests ect on monday xD

Answer 29

oml summer class :o

Answer 30

yerp ;(

Answer 31

aah its a pain x.x im taking calc2 in summer its too much

Answer 32

do they go through it faster in summer? :o
sounds tricky

Answer 33

yesh very fast D: its insane