Question

calculate the indefinite integral of ∫(ln(acos(x)) dx)/(acos(x) √(1-x^2 ))

Answer 1

all yours!

Answer 2

its a setup!

Answer 3

oh so it is

Answer 4

I got it on a homework ;(

Answer 5

\[\int\limits_{}^{}\frac{\ln(a \cos(x) ) dx}{a \cos(x) \sqrt{1-x^2}}\]

Answer 6

http://knowyourmeme.com/memes/its-a-trap#.Tr3W40NHJ4c

Answer 7

im assuming thats arccos.

Answer 8

so i have the problem right problem?

Answer 9

i think it is
\[arc\cos(x)\] rather than
\[acos(x)\] makes it trivialish

Answer 10

oh i thought a was a constant

Answer 11

i did too, but that sqrt at the end sorta screams "setup!!"

Answer 12

yea lol, srry, i ushually use to the negative 1st power

Answer 13

if so try letting u=arccos(x)

Answer 14

i did

Answer 15

i was hoping i missed something. or is my math teacher just being mean?

Answer 16

let u = ln(arccosx), then du would be 1/arccosx . 1/sqrt(1-x^2)

Answer 17

right? or am i getting my derivative of arccos wrong again >.>

Answer 18

because i hate calculus <.<

Answer 19

no it is right and myininaya is about to prove it again i can tell

Answer 20

i think its negative

Answer 21

\[\int\limits_{}^{} \frac{\ln(u)}{u} (-du)\]
let v=ln(u) then dv=1/u du

\[-\int\limits_{}^{} v dv\]

Answer 22

is it really that easy
or did i make mistake?

Answer 23

yes is it a setup as joemath said

Answer 24

its a TRAP!

Answer 25

tarp it's a tarp

Answer 26

\[-\frac{v^2}{2}+C=-\frac{[\ln(u)]^2}{2}+C=-\frac{[\arccos(x)]^2}{2}+C\]

Answer 27

now you messed up

Answer 28

oops i forgot the natural log

Answer 29

forgot the log!

Answer 30

\[\frac{-[\ln|\arccos(x)|]^2}{2}+C\]

Answer 31

how satellite i thought i was going to prove something
i'm unpredictable haha

Answer 32

bam.

Answer 33

lolol

Answer 34

lol thanks for helping me maintain my 100% :)

Answer 35

damn. i guess this time it was ok to remember...

Answer 36

lol

Answer 37

this is the part where i told to class i showed this many times so you do it on your own now have fun

Answer 38

averaging less than 4 hours of sleep really isn't good for doing math. for which i blame the IB diploma program

Answer 39

hate calculus >.> hey satellite, for that problem i brought up last night, apparently we were supposed to guess what form the anti-derivative was in, and solve a system of equations to show it was rational >.>

Answer 40

omg i never would have thought that joe for your problem
i looked at too

Answer 41

i dont know how anyone is supposed to guess that, but whatev >.<

Answer 42

all of my sentences are coming out incomplete
its like i don't read before i click post

Answer 43

what? what system of equations?

Answer 44

zarkon deleted is post lol

Answer 45

his*

Answer 46

like, it was basically:\[\int\limits\frac{x^4-2x^3+3x^2-2x+1}{(x^3-3x+1)^2}\]and we are supposed to say "that kinda looks like a quotient rule with a 2nd degree polynomial over a 3 degree polynomial" >.>

Answer 47

i finally win

Answer 48

what did you win? a cookie?

Answer 49

yes give it back please

Answer 50

well it does look like a quotient rule problem because the denominator is something squared

Answer 51

so we make a guess that:\[F(x)=\frac{ax^2+bx+c}{x^3-3x+1}\]take the derivative, and solve for a b and c. it gives a = -1, b = 1, and c = 0.

Answer 52

thats kindof neat
that derivative is nasty though

Answer 53

yeah =/ but it worked lol. now i have to work on making my solution look neater and more understandable.

Answer 54

good lord

Answer 55

hate calculus >.>

Answer 56

yeah it pretty much s*cks i agree.

Answer 57

i can't say differentiate very well
so i say take derivative

Answer 58

why they make such big words
i'm almost blonde come on

Answer 59

almost blond? lol
and the worst thing about calculus is it is endless

Answer 60

i use to not be able to say statistics without stuttering

Answer 61

calc 1
calc 2
calc 3
calc 10
calc theta
etc etc

Answer 62

dont forget pre-calc >.> lolol

Answer 63

i bet you can't say
'stuttering statistics" without stuttering

Answer 64

lol