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Math Help - Integration

  1. #1
    No one in Particular VonNemo19's Avatar
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    Integration

    \int{csc2x}dx

    1. I used the identity csc2x=\frac{1}{2\sin{x}\cos{x}}
    I tried letting u=\sin{x}. So

    \frac{1}{2}\int{\frac{1}{u-u^3}}du.

    I need an explanation from top to bottom guys, please.

    I'm purposefully trying not to use partial fraction decomp.
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  2. #2
    Eater of Worlds
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    \int csc(2x)dx=\int sec(\frac{\pi}{2}-2x)dx

    Now, integrate the sec. Know that one?.
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by galactus View Post
    Know that one?.
    Uhhhh.....

    Let's say I was going to employ decomp then. How would I do it?
    Last edited by VonNemo19; July 24th 2009 at 04:46 PM.
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  4. #4
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    Hello, VonNemo19!

    There is a formula for this problemL . \int\csc\theta\,d\theta \;=\;\ln|\csc\theta - \cot\theta| + C.
    No one has taught it to you?


    \int\csc2x\,dx
    You can derive the formula like this . . .

    We have: . \int \csc\theta\,d\theta

    Multiply by \frac{\csc\theta + \cot\theta}{\csc\theta + \cot\theta}\!:\quad \int\frac{\csc\theta}{1}\cdot\frac{\csc\theta + \cot\theta}{\csc\theta + \cot\theta}\,d\theta . =\;\int\frac{\csc^2\!\theta + \csc\theta\cot\theta}{\csc\theta + \cot\theta}\,d\theta


    Let u \:=\:\csc\theta + \cot\theta \quad\Rightarrow\quad du \:=\:(-\csc\theta\cot\theta - \csc^2\!\theta)\,d\theta

    . . Hence: . (\csc^2\!\theta + \csc\theta\cot\theta)\,d\theta \:=\:-du

    Substitute: . -\int\frac{du}{u} \;=\;-\ln|u| + C


    Back-substitute: . -\ln|\csc\theta + \cot\theta| + C \;=\; \ln|\csc\theta + \cot\theta|^{-1} + C \;=\; \ln\left|\frac{1}{\csc\theta + \cot\theta}\right| + C

    Multiply by \frac{\csc\theta - \cot\theta}{\csc\theta - \cot\theta}\!:\quad \ln\left|\frac{1}{\csc\theta + \cot\theta}\cdot\frac{\csc\theta - \cot\theta}{\csc\theta-\cot\theta}\right| + C . = \;\ln\left|\frac{\csc\theta - \cot\theta}{\csc^2\!\theta - \cot^2\!\theta}\right| + C


    Since \csc^2\!\theta-\cot^2\!\theta \:=\:1, we have: . \ln|\csc\theta - \cot\theta| + C


    Now apply this formula to your problem . . .

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  5. #5
    Member McScruffy's Avatar
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    Multiply by \frac{\csc\theta + \cot\theta}{\csc\theta + \cot\theta}\!:\quad \int\frac{\csc\theta}{1}\cdot\frac{\csc\theta + \cot\theta}{\csc\theta + \cot\theta}\,d\theta
    This is the approach that I have seen, but can someone explain why it is the most ideal as it seems rather non-intuitive to me.
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  6. #6
    Eater of Worlds
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    Here is another way. More involved.

    \int csc(2x)dx

    \int\frac{1}{sin(2x)}dx

    \int\frac{1}{sin(2x)}\cdot\frac{sin(2x)}{sin(2x)}d  x

    \int\frac{sin^{2}(2x)}{1-cos^{2}(2x)}dx

    Let u=cos(2x), \;\ \frac{-du}{2}=sin(2x)dx

    This gives:

    \frac{-1}{2}\int\frac{1}{1-u^{2}}du

    \frac{1}{4}\int\frac{1}{u-1}du-\frac{1}{4}\int\frac{u+1}du

    \frac{ln(u-1)}{4}-\frac{ln(u+1)}{4}

    \frac{1}{4}ln(\frac{u-1}{u+1})

    Resub:

    \frac{1}{4}ln(\frac{cos(2x)-1}{cos(2x)+1})

    Inside the parentheses, multiply top and bottom by cos(2x)+1

    (one can also use cos(2x)-1)

    This gives:

    \frac{1}{4}ln(\frac{sin^{2}(2x)}{(cos(2x)+1)^{2}})

    Without further detail, the identity inside the parentheses is equal to tan^{2}(x)

    and we get:

    \frac{1}{4}ln(tan^{2}(x))

    By the property of logs:

    \boxed{\frac{1}{2}ln(tan(x))}
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  7. #7
    No one in Particular VonNemo19's Avatar
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    Could I do it this way?

    \int\csc{2x}dx=\int\frac{1}{2\sin{x}\cos{x}}dx

    Let u=\sin{x}, then du=\cos{x}dx

    \int{\frac{1}{2\sin{x}\cos{x}}}dx=\int\frac{1}{2u\  cos^2{x}}du =\int\frac{1}{2u(1-\sin^2{x})}du=\int\frac{1}{2u(1-u^2)}du=\int\frac{1}{2u(1-u)(1+u)}du

    But, \frac{1}{2u(1-u)(1+u)}=\frac{A}{2u}+\frac{B}{1-u}+\frac{C}{1+u}

    1=A(u-1)(u+1)+B2u(u+1)+C2u(1-u)\Rightarrow{A}=-1,B=\frac{1}{4},C=-\frac{1}{4}

    Then \int\csc2xdx=-\frac{1}{2}\ln{|u|}+\frac{1}{4}\ln{|1-u|}-\frac{1}{4}\ln{|1+u|}+C

    If I sub back for u...
    Last edited by VonNemo19; July 24th 2009 at 08:30 PM. Reason: fixed signs
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  8. #8
    Member McScruffy's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    1=A(u-1)(u+1)+B2u(u+1)+C2u(1-u)\Rightarrow{A}=-1,B=\frac{1}{4},C=\frac{1}{4}
    Shouldn't that be 1=A(1-u)(1+u)+B2u(1+u)+C2u(1-u) which gives A=1, B=\frac {1}{4}, C=\frac{-1}{4}? At least that's my thought.
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  9. #9
    Senior Member DeMath's Avatar
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    Or

    \int {\csc 2xdx}  = \int {\frac{{dx}}<br />
{{\sin 2x}}}  = \frac{1}<br />
{2}\int {\frac{{dx}}<br />
{{\sin x\cos x}}}  = \frac{1}<br />
{2}\int {\frac{{{{\sin }^2}x + {{\cos }^2}x}}<br />
{{\sin x\cos x}}dx}  =

    = \frac{1}<br />
{2}\int {\frac{{\sin x}}<br />
{{\cos x}}} dx + \frac{1}<br />
{2}\int {\frac{{\cos x}}<br />
{{\sin x}}} dx =  - \frac{1}<br />
{2}\int {\frac{{d\left( {\cos x} \right)}}<br />
{{\cos x}}}  + \frac{1}<br />
{2}\int {\frac{{d\left( {\sin x} \right)}}<br />
{{\sin x}}}  =

     =  - \frac{1}<br />
{2}\ln \left| {\cos x} \right| + \frac{1}<br />
{2}\ln \left| {\sin x} \right| + C = \frac{1}<br />
{2}\ln \left| {\tan x} \right| + C.
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  10. #10
    No one in Particular VonNemo19's Avatar
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    Can anyone verify my last post?
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  11. #11
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    your last post looks good, but I'd highly recommend memorizing the general csc integration formula that Soroban posted
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  12. #12
    Super Member Random Variable's Avatar
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     \int \csc(2x) \ dx = \int \frac{dx}{\sin(2x)} = \int \frac{1+\tan^{2}(x)}{2\tan(x)} \ dx

    let u = \tan(x)

    then  x = \arctan(u)

    and dx = \frac{du}{1+u^{2}}

     = \int \frac{1+u^{2}}{2u} \frac{du}{1+u^{2}} = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2}ln(|u|) + C = \frac{1}{2}\ln(|\tan(x)|) + C
    Last edited by Random Variable; July 25th 2009 at 07:53 PM.
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  13. #13
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by artvandalay11 View Post
    ...but I'd highly recommend memorizing the general csc integration formula that Soroban posted
    Why? There clearly is abundance of simpler ones. Including the one I derived. It involves only sine. Is there a good reason why soroban's is the best?
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  14. #14
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    Quote Originally Posted by VonNemo19 View Post
    Why? There clearly is abundance of simpler ones. Including the one I derived. It involves only sine. Is there a good reason why soroban's is the best?
    Soroban's identity is elegant, and standard. It's good to know the integration formula for every trigonometric function. More importantly, the solution you derived was for a different integral, and it only (directly) involves the sine function because there was a double angle involved. In other words, he was suggesting you derive your result from the formula for

    \int\csc x\, dx,

    whereas you derived from scatch a formula for

    \int \csc 2x\, dx,

    which is somewhat different.

    Incidentally, and apropos of nothing, do you know the method of Weierstrass substitution? It transforms any rational function in terms of sine and cosine into an ordinary rational function in a single variable.
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  15. #15
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by AlephZero View Post
    More importantly, the solution you derived was for a different integral, and it only (directly) involves the sine function because there was a double angle involved.
    Indeed! I can't believe I missed that. Thank you.

    Quote Originally Posted by AlephZero View Post
    Incidentally, and apropos of nothing, do you know the method of Weierstrass substitution?
    No, but I vaguely recall writing trig functions as algebraic functions; or was that inverse trig functions...?

    I can't remember. I do know that the domain would have to be restricted...
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