Question

is there another method to integrate 1/(5 - 4 cos x) without using the substitution t= tan x/2 ? which doesnt involve all of this http://assets.openstudy.com/updates/attachments/52bbcb52e4b01cdca44a1389-fibonaccichick666-1388041246204-open.png ?

Answer 1

i know its something really simple but i just cant remember the trick...

Answer 2

We would ummmmmm....

Multiply the numerator and denominator by the `conjugate` of the denominator.
That should simplify things down a bit.

Answer 3

\[\Large\bf\sf \int\limits \frac{1}{5-4\cos x}\left(\frac{5+4\cos x}{5+4\cos x}\right)\;dx\]

Answer 4

Hmmm I wonder if that actually helps +_+ thinking....
Cause we can't use our nice square identity in the denominator...

Answer 5

conjugate then trig sub?

Answer 6

i cant do anything with the denom.. \[5^{2} - 4^{^{2}} \cos ^{2}x\]

Answer 7

Ya I guess you could do a trig sub :) lol

\[\Large\bf\sf 4\cos x\quad=\quad 5\sin \theta\]Or something maybe :o hmm

Answer 8

Giving you something like\[\Large 5^2-5^2 \sin^2\theta\]in the denominator, yes?

Answer 9

ok, seems like we're getting somewhere... but then what?

Answer 10

I'd do conjugate, then sub (1-\(sin^2 x\)) for \(cos^2 x\)

Answer 11

so you have:
\[\int \frac{5-4cosx}{25-16(1+sin^2 x)}dx\]
now simplify

Answer 12

then do a u substitution. btw sorry I stole your thunder @zepdrix

Answer 13

Well if we're making a substitution, I don't think we want to call our new variable x. :o

Answer 14

Which will run us into a bunch of trouble with replacing the differential dx.
Hmmmmm I'm wondering if this is the right path +_+

Answer 15

What class is this for? Complex analysis?

Answer 16

nooo this is basic calc II guys

Answer 17

substitute sin x = t ? therefore cos x . dx = dt? is that what you mean that would give me \[\frac{ 5- 4 cosx }{ 9- t ^{2}} dx\]

Answer 18

how will i turn that cos x into cos x .dx , and thus get dt so that i can integrate it?

Answer 19

my substitution was wrong that should be a minus sin squared not plus sorry

Answer 20

so what is the derivative of \[9-16sin^2 x\]

Answer 21

\[-32sinx.cosx\] ?

Answer 22

that is correct, but I just realized my plan doesn't work. Give me a second to write this down

Answer 23

ok :D

Answer 24

oh wait i think i just got it , could i go from \[\int\limits_{}^{}\frac{ 5+4cosx }{ 9-t ^{2} } dx\]
to
\[\int\limits_{}^{ }\frac{ 5 }{9-t ^{2} } dx + 4\int\limits_{}^{}\frac{ \cos x }{ 9-t ^{2} } dx\]

finally giving me \[4\int\limits_{}^{}\frac{ 1 }{ 9-t ^{2} } dt\]

as cos x .dx = dt
and just integrate them separately ?
is that it?

Answer 25

@zepdrix ?

Answer 26

wow...no

just use

\[t=\tan\left(\frac{x}{2}\right)\]

Answer 27

no, that doesn't work. Zarkon's works but it's a nasty bit of algebra and i was trying to avoid it

Answer 28

how can i use t=tan(x/2) ?

Answer 29

http://en.wikipedia.org/wiki/Tangent_half-angle_substitution

Answer 30

this is the "step by step"(sort of). as you can see it requires like a page and a half if you show all of the simplifications

Answer 31

oh darn it!!! now i get it.... my teacher keeps using that substitution t= tan x/2 but i just dont seem to get it.... i try to avoid it.. i guess i'll have to get used to it... thanks a ton peeps!

Answer 32

yea, usually I won't just give answers, but we confused you enough for one question. the subs are important though