Question

How are you going to integrate f(x) = integral of -(sin(y)) dy/ (3tany-2cosy) ?

Answer 1

i simplified it into integral of (-sinycosy)dy / (2sin^y +3 siny-2)

Answer 2

Not sure on the steps..but here is the solution its pretty long

http://www.numberempire.com/integralcalculator.php?function=%28-sin%28x%29%29%2F%283*tan%28x%29-2*cos%28x%29%29&var=x&answers=&__utma=164170016.165112442.1306893325.1312516703.1312708626.8&__utmz=164170016.1312708628.8.7.utmcsr%3Dgoogle |utmccn%3D%28organic%29|utmcmd%3Dorganic|utmctr%3Dintegral+calculator&screen_height=768&__utmb=164170016.1.10.1312708628&__utmc=164170016

Answer 3

Wolfram has some good steps for the integral, I am just wondering if it is possible to get a sensible trig expression from the original...

Answer 4

I think we can....

Answer 5

i can simplify it to this...then maybe partial fractions ?
\[-\frac{\sin x *\cos x}{(2\sin x +1)(\sin x -2)}\]

Answer 6

Think I something better, maybe u check me, divide original top and bottom by 2 siny.

Answer 7

ok

Answer 8

Might have messed it up (not awake yet:-) but

-2Tan^2y/(3Tan^2y-2)

Answer 9

wait how do you get tan^2 ?
divide top by 2sin --> -1/2
divide bottom by 2sin --> (3/2cos) - cot

Answer 10

Ooops, told u not awake:-)

Still, I thing something is there...

Answer 11

Think..

Answer 12

ok
put it together...
\[\frac{\tan x}{3\sin x - 2}\]

Answer 13

That is already easier, right..?

Answer 14

i dunno, what can we substitute here ??

Answer 15

Maybe something /u^2+1...

Answer 16

http://www.wolframalpha.com/input/?i=int%28-sin%28y%29%2F%283tan%28y%29-2cos%28y%29%29%29&a=*C.int-_*IntegralsWord-

Answer 17

i did not see to do this

Answer 18

wolfram will show you the steps

Answer 19

they do u=tan(y/2)
and then use partial fractions

Answer 20

It is the partial fractions I am trying to get out of.....

I'm sure there is a way, just haven't seen it yet...:-)

Answer 21

Maybe try to use sec instead...

Answer 22

eewww calculus.

Answer 23

Well, trig at the moment....

Answer 24

actually this problem does look ewwwie

Answer 25

ewwwie - another one for me to learn..:-)?

Answer 26

urban dictionary says it is an ex of some kind...

Answer 27

ok i just worked it out using the simplified expression i posted above...partial fractions does work
u = sin(x) - 2
du = cos(x)

Answer 28

There u go, I don't like partial fractions though, evil...

Answer 29

haha :)

Answer 30

my steps aren't as messy as wolframs...lol

Answer 31

i don't like using sec and tan if i can help it, just stick with sin and cos

Answer 32

manoranjan just told me in chat that eewwie is dental floss..

Answer 33

yeah i'm sure there are a lot of things that can be defined as "eewwie",

Answer 34

Or ewwwie..

Answer 35

so wheres "thebeefbowl" ?
nice name btw
hopefully all this will be of some help, i don;t think i want to write out my whole solution

Answer 36

Sure he can do with this info...

Answer 37

Ok

\[\frac{-siny}{3tany -2cosy}dy\]
\[Put\:\:\:\:\:\:\: t = siny \]\[\frac {dt}{cosy} = dy\]
\[\frac{-t}{3t -2 + 2t^2}dt\]
Now, \[ \lambda \frac{d(3t -2 + 2t^2)}{dt} + \mu = -t\]

Answer 38

Ooooh....!

Answer 39

Course, the trig would be easy if it wasn't for the 3 and the 2....

Answer 40

Using this we Have
\[3\lambda + 4t\lambda +\mu = -t\]Compare coefficients
\[3\lambda + \mu = 0\]
\[4\lambda = -1\]

Answer 41

G, epsilons next...:-)

Answer 42

Now when we Integrate it
\[f(t) = 3t - 2 +2t^2\]

\[\frac{\lambda f'(t) + \mu }{f(t)} dt\]

lol@estudier I am just trying to do it thats all

Answer 43

No, no...I like it, keep going...:-)

Answer 44

\[\lambda \frac{f'(t)}{f(t)} dt + \frac{\mu}{f(t)}dt\]
Now first part integral is

\[\lambda log(f(t))\]

Answer 45

I keep getting kicked off...

Answer 46

Now all we have to do is the second part

\[\mu \frac{1}{3t - 2 + 2t^2} dt \]

Answer 47

\[3t - 2 + 2t^2 = ((t + 2) (2t -1)\]

\[ \frac{1}{3t - 2 +2t^2} = \frac{A(2t -1) +B(t +2)}{3t - 2 + 2t^2 } \]

Answer 48

Now, Compare coefficients

The Answer should be

\[\lambda log(3t + 2t^2 -2 ) + \mu A log(t+2) + \mu B log(2t -1)\]

Answer 49

ah I am not putting coefficients or t ...you should do algebra on you own

Answer 50

I think I did wrong ...wolfram is showing something very complex

Answer 51

Don't pay attention to Wolfram on this one....

Answer 52

Maybe your right

Answer 53

-sin2y /2 (siny+2)(2siny-1)....Best I can get, I think.

Answer 54

That's from multiplying original integral by cosy top and bottom and u can factor the bottom as a quadratic in sin y...

Answer 55

Mind u I just noticed in the originall question he wants to integrate the integral,lol

Answer 56

lol

Answer 57

haha thanks everyone. i got the idea now. and my question is wrong. it should be the integral o f the function only not the integral of integral :D

Answer 58

my answer is 1/10 ln|2sinx -1| + 2/5 ln |sinx+2| +C