Question

Integrate (cosxsinx) / (cos^2(x) - 1)

Answer 1

\[\int\limits_{}^{}\frac{ cosxsinx }{ \cos^2x -1 }dx\]

Answer 2

u can write cos^2 x as 1-sin^2 x

Answer 3

ok\[\frac{\sin x \cos x}{\cos^2 x-1}=\frac{1}{2}\frac{\sin 2x }{\frac{1+\cos 2x}{2}-1}\]

Answer 4

maybe no need to go for 2x angle.

Answer 5

let u = cos^2 x - 1

du = 2cosx sinx

it's much easier...

Answer 6

sinxcosx/(-sin^2x)

Answer 7

-cotx

Answer 8

lol...mine is much harder

Answer 9

\[\implies \frac 12 \int \frac 1u du\]

Answer 10

then u can cancel sin x from numerator and denominator to get -cot x

Answer 11

i'll organize my solution...

\[\int \frac{\cos x \sin x}{\cos^2 x - 1}dx\]
let \[u = \cos^2 x - 1\]
\[du = -2\sin x \cos x dx\]

so the integral becomes
\[\implies -\frac 12 \int \frac {du}u\]

Answer 12

integrate that

Answer 13

let (cosx)^2 = t

2(cosx)(-sinx)dx = dt

so given eq is integral[1/2(1/(1-t^2)]dt

Answer 14

if you let cos^2 x = t

then your denominator should be t - 1 @LeeYeongKyu

Answer 15

no - is calculated in dominater.

Answer 16

or if you factor out negative...should be 1-t

Answer 17

(cosx)^2 = -2cosxsinx

Answer 18

if you differentite it

Answer 19

you let cos^2 x = t

so your denominator should be 1 - t

Answer 20

sorry... i miss wrote..
thanx

Answer 21

@LeeYeongKyu
http://i.imgur.com/AaCa1.gif

Answer 22

i had a little hasitate

Answer 23

here's another solution..

\[\int \frac{\cos x \sin x}{\cos^2 x - 1}dx\]
let \[u = \cos x\]
\[du = -\sin xdx\]

so integral becomes..
\[\implies -\int \frac{udu}{u^2 - 1}\]
it's harder..bot possible

Answer 24

but*

Answer 25

no it's easy thing to integrate...

Answer 26

i meant harder than what i first wrote

Answer 27

right

Answer 28

the answer I've been getting is \[\frac{ 1 }{ 2}\ln \left| \cos^2x-1\right| + c\], the correct answer according to the book is \[\ln \left|cscx\right| + c\]

Answer 29

\[\frac 12 \ln | \cos^2x - 1| + c \implies \frac 12 \ln |-\sin^2 x| + c\]
\[\implies \frac 12 \ln | \csc^2 x| + c\]
\[\implies \ln |\csc x| + c\]
did you follow that?

Answer 30

so \[\cos^2x-1 = -\sin^2x = \csc^2x\]?

Answer 31

nevermind that..

Answer 32

\[\cos^2 x - 1 = -(1-\cos^2 x) = -\sin ^2 x\]
right?

Answer 33

I'm really bad at my trig identities, if you say so I'll believe it

Answer 34

this came from \[\sin^2 x + \cos ^2 x = 1\]
familiar?

Answer 35

yes

Answer 36

ok I see it now

Answer 37

wonderful

Answer 38

so you have \[\frac 12 \ln | -\sin^2 x| + c\]

right?

Answer 39

right

Answer 40

this is why i told you that you misplaced a negative sign... in this case..the absolute value will just cancel out the negative sign...so you can't get csc^2 x

Answer 41

your integral should have been \[-\frac 12 \ln | \cos^2 x - 1| + c\]
because this gives you \[-\frac 12 \ln | \sin ^2 x| + c\]
if you use power rule here...
\[\implies \ln |(\sin ^2 x)^{-1/2} | + c\]
that becomes \(\ln |\csc x| + c\]
does that make sense to you?

Answer 42

or should i try a different approach to make you see what i mean?

Answer 43

ok I see, you used clnx = lnx^c so that we could get 1/sin^2x which equals cscx

Answer 44

right!

Answer 45

:D thanks so much for sticking it out to the end!

Answer 46

welcome