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Math Help - Possibly a minor puzzle

  1. #1
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    Possibly a minor puzzle

    One of my interests is integral calculus so when I conceived of a problem such as

    \int dx/(1 + cosx)(1 + sinx)

    that resists methods such as integration by parts or a substitution, this gets me interested as it's so far been challenging to me.

    Even though I feel that a simple (closed) solution exists for this problem, I haven't found one to date so I figure that others may want to take a look and try their hand at it.

    Any ideas?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by wonderboy1953 View Post
    Any ideas?

    t=\tan \dfrac {x}{2}


    Fernando Revilla
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  3. #3
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by FernandoRevilla View Post
    t=\tan \dfrac {x}{2}


    Fernando Revilla
    That is a beast: dx/ ( (1+cosx)(1+sinx)) - Wolfram|Alpha
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  4. #4
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    Hello, wonderboy1953!

    Actually, Fernando's suggestion isn't that bad.
    And I don't understand Wolfram's complex approach.


    \displaystyle\int \frac{dx}{(1 + \cos x)(1 + \sin x)}

    From t = \tan\frac{x}{2} we have: . \begin{Bmatrix}dx &=& \dfrac{2\,dt}{1+t^2} \\ \\[-3mm] \sin x &=& \dfrac{2t}{1+z^2} \\ \\[-3mm] \cos x &=& \dfrac{1-t^2}{1+t^2} \end{Bmatrix}

    \displaystyle \text{Substitute: }\;\int \frac{\dfrac{2\,dt}{1+t^2}}{\left(1 + \dfrac{1-t^2}{1+t^2}\right)\left(1 + \dfrac{2t}{1+t^2}\right)}

    . . . . . . . \displaystyle =\;\int\frac{\dfrac{2\,dt}{1+t^2}}{\left(\dfrac{1+  t^2 + 1 - t^2}{1+t^2}\right)\left(\dfrac{1+t^2+2t}{1+t^2}\ri  ght)}

    . . . . . . . \displaystyle =\;\int\frac{\dfrac{2\,dt}{1+t^2}}{\left(\dfrac{2}  {1+t^2}\right)\dfrac{(1+t)^2}{1+t^2}}


    \displaystyle\text{Multiply by }\tfrac{(1+t^2)^2}{(1+t^2)^2}\!:\;\;\;\int \frac{2(1+t^2)\,dx}{2(1+t)^2} \;=\;\int\frac{1+t^2}{(1+t)^2}\,dt

    . . \displaystyle =\;\int\frac{1 + {\bf2t} + t^2 - {\bf2t}}{1+2t+t^2}\,dt \;=\;\int\left(1 - \frac{2t}{1+2t+t^2}\right)\,dt

    . . \displaystyle =\;\int\left(1 - \frac{2t + {\bf2} - {\bf2}}{1+2t+t^2}\right)\,dt \;=\; \int\left(1 - \frac{2t+2}{1+2t+t^2} + \frac{2}{(1+t)^2}\right)\,dt

    . . \displaystyle =\;\int\left(1 - \frac{2 + 2t}{1+2t+t^2} + 2(1+t)^{-2}\right)\,dt


    . . =\; t - \ln(1+t)^2 - 2(1+t)^{-1} + C


    . . =\;t - 2\ln|1+t| - \dfrac{2}{1+t} + C


    Back-substitute:

    . . \tan\frac{x}{2} - \ln\left|1 + \tan\frac{x}{2}\right| - \dfrac{2}{1 + \tan\frac{x}{2}} + C

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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by AllanCuz View Post
    That is a beast
    That is not a quantivative question but qualitative, the substitution t=\tan (x/2) is "decidable" for all \int R(\sin x,\cos x)\;dx whith R rational.


    Fernando Revilla
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  6. #6
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    Thank you

    I say that Fernando Revilla wins out on this puzzle although I'd say we all win out from our knowledge gained.

    I've been promoting what I refer to as the A-S method (the addition-subtraction method) for use in helping to solve certain integration problems (this method was inspired by what I read on LaPlace transforms from Rainville-Bedient's Elementary Differential Equations) as it simplifies those problems' solutions. I tried the A-S method on the puzzle which failed to work so Fernando Revilla referred to another method that, while it works and is listed in Schaum's outline, isn't simple to apply (as Schaum demonstrates).

    I will continue my integration and differential equations studies to see what else may be learned.
    Last edited by wonderboy1953; January 30th 2011 at 11:07 AM.
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  7. #7
    Math Engineering Student
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    here's my approach:

    first we calculate some integrals separately:

    \displaystyle\int{\frac{dx}{1+\sin x}}=\int{\frac{1-\sin x}{{{\cos }^{2}}x}\,dx}=\tan (x)-\sec (x)+{{k}_{1}},\, (1).

    \displaystyle\int{\frac{1-\sin x}{{{\sin }^{2}}x}\,dx}=-\cot (x)-\ln \left| \csc (x)-\cot (x) \right|+{{k}_{2}},\, (2).

    \begin{aligned}<br />
   \int{\frac{\cos x}{(1+\sin x){{\sin }^{2}}x}\,dx}&=\int{\frac{(1-t)(1+t)+{{t}^{2}}}{(1+t){{t}^{2}}}\,dt},\text{ }t=\sin x \\ <br />
 & =-\frac{1}{t}-\ln \left| t \right|+\ln \left| 1+t \right|+{{k}_{3}} \\ <br />
 & =\ln \left| \frac{1+\sin x}{\sin x} \right|-\frac{1}{\sin x}+{{k}_{3}},\,(3) \\ <br />
\end{aligned}

    now we're ready to solve your integral, so by using (1), (2) and (3) we get

    \begin{aligned}<br />
   \int{\frac{dx}{(1+\cos x)(1+\sin x)}}&=\int{\frac{1-\cos x}{(1+\sin x){{\sin }^{2}}x}\,dx} \\ <br />
 & =\int{\frac{{{\sin }^{2}}x+{{\cos }^{2}}x}{(1+\sin x){{\sin }^{2}}x}\,dx}-\int{\frac{\cos x}{(1+\sin x){{\sin }^{2}}x}\,dx} \\ <br />
 & =\tan (x)-\sec (x)-\cot (x)-\ln \left| \csc (x)-\cot (x) \right| \\ <br />
  &-\ln \left| \frac{1+\sin x}{\sin x} \right|+\csc (x)+k. \\ <br />
\end{aligned}
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  8. #8
    Math Engineering Student
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    i just found a better solution,

    \begin{aligned}<br />
   \int{\frac{dx}{(1+\cos x)(1+\sin x)}}&=\int{\frac{(1-\cos x)(1+\cos x)+(1-\sin x)(1+\sin x)}{(1+\cos x)(1+\sin x)}\,dx} \\ <br />
 & =\int{\frac{1-\cos x}{1+\sin x}\,dx}+\int{\frac{1-\sin x}{1+\cos x}\,dx} \\ <br />
 & =\tan (x)-\sec (x)-\cot (x)+\csc (x)+\ln \left| \frac{1+\cos x}{1+\sin x} \right|+k. \\ <br />
\end{aligned}
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  9. #9
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    Not just better

    Quote Originally Posted by Krizalid View Post
    i just found a better solution,

    \begin{aligned}<br />
   \int{\frac{dx}{(1+\cos x)(1+\sin x)}}&=\int{\frac{(1-\cos x)(1+\cos x)+(1-\sin x)(1+\sin x)}{(1+\cos x)(1+\sin x)}\,dx} \\ <br />
 & =\int{\frac{1-\cos x}{1+\sin x}\,dx}+\int{\frac{1-\sin x}{1+\cos x}\,dx} \\ <br />
 & =\tan (x)-\sec (x)-\cot (x)+\csc (x)+\ln \left| \frac{1+\cos x}{1+\sin x} \right|+k. \\ <br />
\end{aligned}
    The best method, no doubt.
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