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Integration of ##1/\sin 2x##

  1. Jul 1, 2016 #1
    1. The problem statement, all variables and given/known data
    ##∫ (1/sin 2x)dx##


    2. Relevant equations


    3. The attempt at a solution
    let ##u = sin 2x, ⇒ du= 2cos2x dx##
    or
    ##sin 2x= 2 sin x cos x##...


    or
    ##∫ (1/sin 2x)dx = ∫( csc 2x)dx##
     
  2. jcsd
  3. Jul 1, 2016 #2

    Math_QED

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    First, substitute the 2x. Then, notice that 1/sinx = sinx/sin^2x.
     
  4. Jul 1, 2016 #3
    i don't get how is ## 1/sin 2x = sin x/sin 2x##? and what do you mean by saying substitute 2x?substitute where?
     
  5. Jul 1, 2016 #4
    i have seen it........## 1/2∫cos u du ##= ( -1/2) ln cot x ## why is textbook answer saying tan x?
     
    Last edited: Jul 1, 2016
  6. Jul 1, 2016 #5

    SammyS

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    I would suppose that Math_QED was suggesting that you use u substitution with u = 2x . That does result in ##\ \frac12 \int \csc(u)\ du\ .##

    However, I would make a different suggestion.
    Simply use the double angle formula for sin(2x) to give 2sin(x)cos(x) .

    You get ##\displaystyle \ \frac12 \int \frac1{\sin(x)\cos(x)}\ dx\ ../ ## Then multiply by either ##\displaystyle\ \frac{\sin(x)}{\sin(x)} \ ## or ##\displaystyle\ \frac{\cos(x)}{\cos(x)} \ .\ ##

    Either will work.
     
  7. Jul 1, 2016 #6
    I got an idea! Try

    1= sin^2 x + cos^2 x and divide the fraction in two parts :)
     
  8. Jul 1, 2016 #7

    SammyS

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    Your question is not clear.

    Do you mean that Math_QED's suggestion gives you ##\ 1/2∫\cos u\ du \ ?\ ## If so, then no, that results in ##\ 1/(2∫(1/\sin u)\ du) \ .\ ##

    Furthermore, ## ( -1/2) \ln\, (\cot x) ## is indeed equal to ## ( 1/2) \ln\, (\tan x) ##

    (Alex has a rather interesting idea !
    Added in Edit:
    Alex's idea doesn't work out the way I at first thought, but maybe it's still helpful, maybe not.)​
     
    Last edited: Jul 1, 2016
  9. Jul 1, 2016 #8
    S (sin^2 x + cos^2 x)/ sin(2x) dx= 1/2 Stanx dx + 1/2 Scotx dx

    Then just integrate tangent and co tangent
     
  10. Jul 1, 2016 #9

    SammyS

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    I assume you use S for ∫ .

    OK!

    Yes, that is very cleaver !
     
  11. Jul 1, 2016 #10

    LCKurtz

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    @Cozma Alex 's answer gives another way to do ## \int \csc x~dx## that I have never seen before. The usual way is to multiply the integrand by ##\frac{\csc x + \cot x}{\csc x + \cot x}## and let ##u = \csc x + \cot x## to get a ##-\frac {du} u## form. I have never liked this method because if you can remember what to multiply by, you might as well just remember the answer. So now, thinking of Cozma's method, if you need to evaluate ## \int \csc x~dx## just let ##x = 2u## to get$$
    \int \csc x~dx = 2\int \csc 2u~du = 2 \int \frac 1 {\sin 2u}~du$$Then use Cozma's method in post #8. Admittedly it is a bit of work, but I like it :-).
     
  12. Jul 1, 2016 #11

    SammyS

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    Yes. I agree!

    As far as I can see, this doesn't work so nicely for integrating sec(x). The only way I could get that integral to work in a similar fashion was to use the co-function identity along with u-substitution.

    ##\displaystyle \int \sec x \,dx=\int \csc \left(\frac{\pi}{2}-x\right) \,dx ##
     
  13. Jul 2, 2016 #12
    nice discussion...looks like there are different ways of doing it...i would vouch for the double angle way...
     
  14. Jul 5, 2016 #13
    I would just use the identity $$1/sinx=cscx$$
     
  15. Jul 5, 2016 #14
    Thanks Midget any particular reason?
     
  16. Jul 5, 2016 #15
    the integration of that identity is $$ln|cscx-cotx|$$.
    There are different proofs for it you can do. The nicest one is using an identity to rewrite cscx, then replace it in the integral. The longer way, without knowing the trick, is to do parts.
     
  17. Jul 5, 2016 #16

    SammyS

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    By the way, since Alex's method works so nicely, I didn't follow up on this.

    Saying to multiply by either ##\displaystyle\ \frac{\sin(x)}{\sin(x)} \ ## or ##\displaystyle\ \frac{\cos(x)}{\cos(x)} \ ## was a bit misleading although technically correct.

    Rather multiply by either ##\displaystyle\ \frac{\displaystyle 1 }{\ \frac{\sin(x)}{\sin(x)}\ } \ ## or ##\displaystyle\ \frac{\displaystyle 1 }{\frac{\ \cos(x)}{\cos(x)\ }} \ .\ ##

    For the former this gives: ##\displaystyle \ \frac{1}{2} \int \frac1{\sin^2(x)\cot(x)}\ dx\ ##

    ##\displaystyle = \frac{1}{2} \int \frac{\csc^2(x)}{\cot(x)}\ dx\ ##​

    This is easily handled with u-substitution.
     
  18. Jul 5, 2016 #17
    nice one Sammy..
     
  19. Jul 18, 2016 #18

    haruspex

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    It is? I've always done it by multiplying top and bottom by sin(x), converting sin(x)dx to -d cos(x) and writing the denominator as 1-cos2(x). The denominator can then be split using partial fractions to produce two integrals with solutions like ln(1+/-cos(x)).
     
  20. Jul 18, 2016 #19

    LCKurtz

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    Heh. By "the usual way" I mean the way I learned it back in the '50's. :smile:
     
  21. Jul 19, 2016 #20
    Another way
    $$
    \int \frac{dx}{sin(2x)} = \int \frac{dx}{2sinx~cosx} = \int \frac{dx}{2cos^2 x~tanx} = \begin{bmatrix}
    t = tan x\\
    dt = \frac{dx}{cos^2x}
    \end{bmatrix} = \frac{1}{2}\int \frac{dt}{t}= \frac{1}{2}ln|t| + C = \frac{1}{2}ln|tan x| + C
    $$
     
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