Question

solve int dx/sinx with x=2t => t := x/2

Answer 1

\[\int\limits \frac{ dx }{ \sin(x) }\]

Answer 2

maybe writing \(\sin x\) in terms of \(\tan \frac{x}{2}\) :)

Answer 3

there's a particular way my professor solves this, i'll show you, but i don't understand the steps

Answer 4

something like: ln(csc+cot) right?

Answer 5

the answer is ln abs(sint/cost) + k

Answer 6

let x = 2t

therefore, dx = 2 dt\[\int \frac{dx}{sin(x)}\to \int \frac{2~dt}{sin(2t)}\]

Answer 7

\[sin(2t)=2sin(t)~cos(t)\]
\[\int\frac{dt}{sin(t)~cos(t)}\]

Answer 8

\[2\int\limits\frac{ dt }{ 2\sin t cost } = \int\limits \frac{ \sin^2t+\cos^2t }{ sint cost }dt= \int\limits tantdt + \int\limits cottdt= -\int\limits \frac{ dcost }{ cost } + \int\limits \frac{ dsint }{ sint }\]

Answer 9

hmm

Answer 10

lol, 1 = sin^2 + cos^2 so thats a nice step

Answer 11

exactly, i understand everything till the - int

Answer 12

how does int tant dt + int cott dt go to the next step

Answer 13

been stuck a few hours i think

Answer 14

\[\int\frac{dt}{sin(t)~cos(t)}\]
\[\int\frac{sin^2(t)+cos^2(t)}{sin(t)~cos(t)}dt\]
\[\int\frac{sin^2(t)}{sin(t)~cos(t)}+\frac{cos^2(t)}{sin(t)~cos(t)}dt\]
\[\int\frac{sin(t)}{cos(t)}+\frac{cos(t)}{sin(t)}dt\]
\[\int\frac{sin(t)}{cos(t)}dt+\int\frac{cos(t)}{sin(t)}dt\]
\[-ln(cos(t))+ln(sin(t))+k\]

Answer 15

\[\frac{d}{du}[ln(u)]=\frac{u'}{u}\]

Answer 16

since sin and cos are derivatives of each other ....

Answer 17

\[\int\frac{sin(t)}{cos(t)}dt+\int\frac{cos(t)}{sin(t)}dt\]
\[\int\frac{-[cos(t)]'}{cos(t)}dt+\int\frac{[sin(t)]'}{sin(t)}dt\]
\[-ln(cos(t))+ln(sin(t))+k\]

Answer 18

alright..let me try and understand

Answer 19

i don't get it, how did you get to ln?

Answer 20

its one of the basic derivative rules that you developed from those limit h to 0 stuff

Answer 21

or if you already know the e setup\[y=ln(u)\]
\[e^y=u\]
\[y'~e^y = u'\]
\[y' = \frac{u'}{e^y}~:~but e^y = u\]
\[[ln(u)]' = \frac{u'}{u}\]

Answer 22

the antiderivative (integral) of the form: u'/u is ln(u)

Answer 23

i see it now, i seem to have skipped this page

Answer 24

i'll try again

Answer 25

:) good luck

Answer 26

thanks

Answer 27

\[\int\limits\frac{-[\cos(t)]'}{\cos(t)}dt+\int\limits\frac{[\sin(t)]'}{\sin(t)}dt\]

Answer 28

how'd you get to this?

Answer 29

because they're derivatives of each other? what do you mean?

Answer 30

what is the derivative of cos(t)?

what is the derivative of sin(t)?

Answer 31

of cost= sint of sint= -cost

Answer 32

but why?

Answer 33

why does that matter?

Answer 34

you have those a little off

[cos(t)]' = -sin(t)
[sin(t)]' = cos(t)

and it matters because we see that the setups relate to the derivative of the ln(u) function. Which of course is why I went all thru the derivative of ln(u) in the first place .....

Answer 35

why does this setup solve into ln(...) + k ? because it matters

Answer 36

@satellite73 any input?

Answer 37

i get it now

Answer 38

because - int dcost/cost lets me use int du/u = ln u + k

Answer 39

which enables me to write it as ln cost

Answer 40

correct

the point is, we have very few formats that we can readily know an integral style to apply. Not all functions can be placed into one of the few formats that can be antiderived. If we can algebra the setup into a familiar format, we can then integrate it up ....

Answer 41

looks like you are doing fine to me. i have never seen it done like this before

Answer 42

feeling great right now

Answer 43

closing this question, and thank you for your help!

Answer 44

good luck ;)

Answer 45

btw, i checked and the derivative of cost=sint and sint=-cost