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Math Help - integrate without complex factorisation

  1. #1
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    integrate without complex factorisation

    integrate
    1/[1+sin(x)+sin^2 (x)]
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by ayushdadhwal View Post
    integrate
    1/[1+sin(x)+sin^2 (x)]
    Observe that \sin^2 x+\sin x+1 = \left(\sin x+\tfrac{1}{2}\right)^2+\tfrac{3}{4}.

    Also observe that \displaystyle\int\frac{\,dx}{\left(\sin x+\tfrac{1}{2}\right)^2+\tfrac{3}{4}} \sim \int\frac{\,du}{u^2+a^2}

    Can you figure out what substitution is required to complete the integration?
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by ayushdadhwal View Post
    integrate
    1/[1+sin(x)+sin^2 (x)]
    If You have an integral containing only circular functions the substitution t= \tan \frac{x}{2} ever conducts to an integral of a rational function. In particular...

    \displaystyle t= \tan \frac{x}{2}

    \displaystyle \implies d x= 2\ \frac{d t}{1+t^{2}}

    \displaystyle \implies \sin x= \frac{2 t}{1+t^{2}}

    \displaystyle \implies \cos x = \frac{1-t^{2}}{1+t^{2}}

    In the case You have proposed the trasformation is...

    \displaystyle  \int \frac{dx}{1+ \sin x + \sin^{2} x} \implies \int \frac{1+t^{2}}{1+2 t+ 6 t^{2} + 2 t^{3} + t^{4}}\ dt

    Kind regards

    \chi \sigma
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  4. #4
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    Quote Originally Posted by Chris L T521 View Post
    Observe that \sin^2 x+\sin x+1 = \left(\sin x+\tfrac{1}{2}\right)^2+\tfrac{3}{4}.

    Also observe that \displaystyle\int\frac{\,dx}{\left(\sin x+\tfrac{1}{2}\right)^2+\tfrac{3}{4}} \sim \int\frac{\,du}{u^2+a^2}

    Can you figure out what substitution is required to complete the integration?
    Nice idea but it won't fly. dx \neq du ....
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    Sub.  \displaystyle \tan(x/2) = \frac{1-t}{1+t} or

     \displaystyle x = \frac{\pi}{2} - 2\tan^{-1}(t)

     \displaystyle   dx = -\frac{2dt}{1+t^2}

    At the same time

     \displaystyle  \sin(x) = \frac{1-t^2}{1+t^2}

    We have , the integral

     \displaystyle   = - 2  \int \frac{1+t^2}{ (1-t^2)^2 + (1-t^2)(1+t^2) + (1+t^2)^2 }~dt{

     \displaystyle   = -2 \int \frac{t^2+1}{t^4 + 3 }~dt

     \displaystyle   = -(1 + \frac{1}{\sqrt{3} } )  \int \frac{t^2 + \sqrt{3}}{t^4 + 3 }~dt  - ( 1 -  \frac{1}{\sqrt{3}} ) \int \frac{t^2 - \sqrt{3}}{t^4 + 3 }~dt

    Consider

     \displaystyle   \int \frac{t^2 + \sqrt{3}}{t^4 + 3 }~dt  = \int \frac{ 1 + \sqrt{3}/t^2 }{ ( t - \sqrt{3}/t )^2+ 2 \sqrt{3} } ~dt

    Sub.  \displaystyle   t - \sqrt{3}/t = u

     \displaystyle   (1 + \sqrt{3}/t^2 )dt = du

    The integral

     \displaystyle   = \int \frac{du}{u^2 + 2\sqrt{3} } = \frac{1}{\sqrt[4]{12}}\tan^{-1}( \frac{u}{\sqrt[4]{12}}) + C

    How about the second integral ?

     \displaystyle   \int \frac{t^2 - \sqrt{3}}{t^4 + 3 }~dt
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    Quote Originally Posted by simplependulum View Post
    How about the second integral ?

     \displaystyle   \int \frac{t^2 - \sqrt{3}}{t^4 + 3 }~dt
    \begin{aligned} & \displaystyle t^4+3 = (t^2)^2+(\sqrt{3})^2 = (t^2+\sqrt{3})^2-2\sqrt{3}t^2 = (t^2-\sqrt{2}\sqrt{\sqrt{3}}t+\sqrt{3})(t^2+\sqrt{2}\sq  rt{\sqrt{3}}t+\sqrt{3}). \\& \therefore ~ I = \int\frac{t^2-\sqrt{3}}{t^4+3}\;{dt} = \int\frac{t^2-\sqrt{3}}{ (t^2-\sqrt{2}\sqrt{\sqrt{3}}t+\sqrt{3})(t^2+\sqrt{2}\sq  rt{\sqrt{3}}t+\sqrt{3})}\;{dt}.\end{aligned}

    \displaystyle u :=  \frac{t^2-\sqrt{2}\sqrt{\sqrt{3}}t+\sqrt{3}}{t^2+\sqrt{2}\sq  rt{\sqrt{3}}t+\sqrt{3}} \Rightarrow \frac{du}{dt} = \frac{2\sqrt{2}\sqrt{\sqrt{3}}\left(t^2-\sqrt{3}\right)}{(t^2+\sqrt{2}\sqrt{\sqrt{3}}t+\sq  rt{3})^2} \Rightarrow dt = \frac{(t^2+\sqrt{2}\sqrt{\sqrt{3}}t+\sqrt{3})^2}{2  \sqrt{2}\sqrt{\sqrt{3}}\left(t^2-\sqrt{3}\right)}\;{du}.

    \begin{aligned}  \therefore ~ I & =  \int\frac{(t^2-\sqrt{3})(t^2+\sqrt{2}\sqrt{\sqrt{3}}t+\sqrt{3})^2  }{2\sqrt{2}\sqrt{\sqrt{3}}(t^2+\sqrt{2}\sqrt{\sqrt  {3}}t+\sqrt{3})(t^2-\sqrt{2}\sqrt{\sqrt{3}}t+\sqrt{3})\left(t^2-\sqrt{3}\right)}\;{du} \\& = \frac{1}{2\sqrt{2}\sqrt{\sqrt{3}}}\int\frac{(t^2+\  sqrt{2}\sqrt{\sqrt{3}}t+\sqrt{3})}{(t^2-\sqrt{2}\sqrt{\sqrt{3}}t+\sqrt{3})}\;{du} = \frac{1}{2\sqrt{2}\sqrt{\sqrt{3}}}\int\frac{1}{u}\  ;{du} \\& = \frac{1}{2\sqrt{2}\sqrt{\sqrt{3}}}\ln{u}+k = \boxed{\frac{1}{2\sqrt{2}\sqrt{\sqrt{3}}}\ln{\frac  {(t^2-\sqrt{2}\sqrt{\sqrt{3}}t+\sqrt{3})}{(t^2+\sqrt{2}\  sqrt{\sqrt{3}}t+\sqrt{3})}}+k}.\end{aligned}

    I received an offer from one of my university choices while typing this up!
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  8. #8
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    Quote Originally Posted by simplependulum View Post
    How about the second integral ?

     \displaystyle   \int \frac{t^2 - \sqrt{3}}{t^4 + 3 }~dt
    i'd write it as in the same fashion,

    \dfrac{1-\frac{\sqrt{3}}{{{t}^{2}}}}{{{\left( t+\frac{\sqrt{3}}{t} \right)}^{2}}-2\sqrt{3}},

    and substitution t+\dfrac{\sqrt{3}}{t}=\sqrt{2}\sqrt[4]{3}u will solve the problem.
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