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Thread: 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 $\displaystyle \sin^2 x+\sin x+1 = \left(\sin x+\tfrac{1}{2}\right)^2+\tfrac{3}{4}$.

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

    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 $\displaystyle t= \tan \frac{x}{2}$ ever conducts to an integral of a rational function. In particular...

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

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

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

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

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

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

    Kind regards

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

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

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

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

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

    At the same time

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

    We have , the integral

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

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

    $\displaystyle \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 \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 \displaystyle t - \sqrt{3}/t = u $

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

    The integral

    $\displaystyle \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 \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 \displaystyle \int \frac{t^2 - \sqrt{3}}{t^4 + 3 }~dt $
    $\displaystyle \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} $

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

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

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

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