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Math Help - Crazy integral

  1. #1
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    Crazy integral

    Determine which integral is more crazy ...

     \int \frac{dx}{1 + \sin^3{x}}

    and

     \int \frac{dx}{1+\sin^6{x}}

    3 or 6 ? The answer is 3 ! It is very interesting that although 6 is greater than 3 , the integral  \int \frac{dx}{1+\sin^6{x}} is actually easier than the other .

    I suggest to try  \int \frac{dx}{1+\sin^6{x}} first , finally finish the rest
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  2. #2
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    If I were to attempt to find \int\frac1{1+\sin^3 x}\mathrm dx then I'd be crazy. I have attempted to find it, and I am crazy ... now.

    Thanks a lot, simplependulum! You have determined which member is more crazy.
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  3. #3
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    Just use the vastly underrated method of Weierstrass Substitution... then both integrals are easy as pie...
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  4. #4
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    Quote Originally Posted by halbard View Post
    If I were to attempt to find \int\frac1{1+\sin^3 x}\mathrm dx then I'd be crazy. I have attempted to find it, and I am crazy ... now.

    Thanks a lot, simplependulum! You have determined which member is more crazy.

    Yes , people can prove they are crazy or not in this question .


    I was wondering which method you have made use , I first deduced a general solution for  \int \frac{ dx}{ \sin^2{x} + b \sin{x} + c  }   ~~ and if the module of the root  u^2 + bu + c = 0 is 1 , it will become easier to solve !
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  5. #5
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    Are you by chance referring to the identity

    \frac3{1+\sin^3 x}=\dfrac1{1+\sin x}+\frac1{1-\omega_1\sin x}+\frac1{1-\omega_2\sin x}

    where \omega_1, \omega_2 are the complex cube roots of -1, together with the substitution t=\tan{\textstyle\frac12}x for \int\dfrac1{1-a\sin x}\mathrm dx which can be found when a^2\neq 1 using \int\frac1{1-2at+t^2}\mathrm dt=\frac{\arctan\left(\displaystyle\frac{t-a}{\sqrt{1-a^2}}\right)}{\sqrt{1-a^2}}? No, I haven't tried it this way (;->)
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