Question

I needz Help. :3

Answer 1

\[\int\limits \frac{ dx }{ cosx-1 }\]

Answer 2

hmm

Answer 3

did you try the conjugate multiply ...cos(x)+1

Answer 4

Ooo.... I have not.

Answer 5

so it would be\[\int\limits \frac{ cosx+1 }{ \cos^2x-1 }\] which makes the denominator -sin^2x ?

Answer 6

or maybe use one of those properties, half angles

sin^2(x/2) = (1-cos(x)) / 2

maybe replace the 1 - cos(x) with that 2sin^2

Answer 7

Or that first way, you can replace the denominator with -sin^2(x)

Answer 8

But then what. :/
I got stuck. lol

Answer 9

umm..

Answer 10

\[\int\limits\limits \frac{ cosx+1 }{ \cos^2x-1 } = -\int\limits \frac{ \cos(x) + 1 }{ \sin^2(x) }\]

Answer 11

that can break into 2 easier integrals to do

\[-\int\limits \frac{ 1 }{ \sin^2(x) }*dx - \int\limits \frac{ \cos(x) }{ \sin^2(x) }*dx\]

Answer 12

first one is just integral of csc^2(x)

second you can substitute u=sin(x) du=cos(x)dx

Answer 13

I got \[\frac{ 1 }{ \sin x }+\cot x+ C\]

Answer 14

Thanks! I got it

Answer 15

yeah looks good i thi nk,

Answer 16

cool , welcome