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Thread: integrate sinx/sin4x

  1. #1
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    integrate sinx/sin4x

    Integrate
    Code:
    sinx/sin4x
    I exapned sin4x and did some partial fractions stuff, but the answer which I get is not the back answer
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  2. #2
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    Quote Originally Posted by ice_syncer View Post
    Integrate
    Code:
    sinx/sin4x
    I exapned sin4x and did some partial fractions stuff, but the answer which I get is not the back answer
    integrate Sin[x]/Sin[4x] - Wolfram|Alpha

    Click on Show steps.
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    ;D
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    Quote Originally Posted by mr fantastic View Post

    Wolfram's steps look too frightening. I'd rather go for a trigonometric substitution:

    $\displaystyle \displaystyle{\tan\frac{x}{2}=t\Longrightarrow dx=\frac{2}{1+t^2}\,dt\,,\,\,\cos x=\frac{1-t^2}{1+t^2}\,,\,\,\sin x=\frac{2t}{1+t^2}}$ , so with a little trigonometry:

    $\displaystyle \displaystyle{\int \frac{\sin x}{\sin 4x}\,dx=\int\frac{\sin x}{2\sin 2x\cos 2x}\,dx=\int\frac{dx}{4\cos x(2\cos^2x-1)}\,dx}=$

    $\displaystyle \displaystyle{\frac{1}{4}\int\frac{(1+t^2)^2}{(1-t^2)\left[t^2-(3+2\sqrt{2})\right]\left[t^2-(3-2\sqrt{2})\right]}}$ , and we have an integral of a rational function.

    Indeed this still is a nasty integral, but perhaps a little less messier than in the steps of the proposed solution

    in Wolfram's...perhaps.

    Tonio
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    Quote Originally Posted by tonio View Post
    $\displaystyle \int\frac{dx}{4\cos x(2\cos^2x-1)}\,dx}$
    $\displaystyle \displaystyle\frac{1}{\cos x(2{{\cos }^{2}}x-1)}=\frac{\cos x}{(1-{{\sin }^{2}}x)(1-2{{\sin }^{2}}x)},$ substitute $\displaystyle t=\sin t$ and get

    $\displaystyle \begin{aligned}
    \int{\frac{dt}{(1-{{t}^{2}})(1-2{{t}^{2}})}}&=\int{\frac{2(1-{{t}^{2}})-(1-2{{t}^{2}})}{(1-{{t}^{2}})(1-2{{t}^{2}})}\,dt} \\
    & =2\int{\frac{dt}{1-2{{t}^{2}}}}-\int{\frac{dt}{1-{{t}^{2}}}\,dt}, \\
    \end{aligned}$

    kill those in the same fashion.

    -----

    that's one of the biggest reasons on why i don't absolute recommend wolfram for tricky problems, we gotta let our people think.
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    Wolfram also made mess of the partial fractions:

    If letting $\displaystyle u = \sin{x}$ gives $\displaystyle \displaystyle \int\frac{1}{8 u^4-12 u^2+4}\;{du} $, then simply write:

    $\displaystyle \displaystyle 8u^4-12u^2+4 = 8t^2-12t+4 = 4(2t^2-3t+1)[/Math]

    $\displaystyle \displaystyle = 4(2t-1)(t-1) = 4(2u^2-1)(u^2-1) $$$\displaystyle etc and use partial fractions.

    Or, alternatively, observe that $\displaystyle 1 = \frac{1}{2}\left\{(u+1)-(u-1)\right\} $, then we have:

    $\displaystyle I = \displaystyle \int\frac{1}{8 u^4-12 u^2+4}\;{du} = \frac{1}{8}\int\frac{(u+1)-(u-1)}{(u-1)(u+1)(2u^2-1)}\;{du}$$$\displaystyle

    [Math]\displaystyle = \frac{1}{8}\int\frac{1}{(u-1)(2u^2-1)}\;{du}-\frac{1}{8}\int\frac{1}{(u+1)(2u^2-1)}\;{du}$

    $\displaystyle \displaystyle = \frac{1}{8}\int\frac{1}{(u-1)(\sqrt{2}u-1)(\sqrt{2}u+1)}\;{du}-\frac{1}{8}\int\frac{1}{(u+1)(\sqrt{2}u-1)(\sqrt{2}u+1)}\;{du}[/tex]

    Observe that $\displaystyle 1 = \frac{1}{2}\left\{(\sqrt{2}u+1)-(\sqrt{2}u-1)\right\}$, then we have:

    $\displaystyle \displaystyle I = \frac{1}{16}\int\frac{(\sqrt{2}u+1)-(\sqrt{2}u-1)}{(u-1)(\sqrt{2}u-1)(\sqrt{2}u+1)}\;{du}-\frac{1}{16}\int\frac{(\sqrt{2}u+1)-(\sqrt{2}u-1)}{(u+1)(\sqrt{2}u-1)(\sqrt{2}u+1)}\;{du}$

    $\displaystyle \displaystyle = \underbrace{\frac{1}{16}\int\frac{1}{(u-1)(\sqrt{2}u-1)}\;{du}}_{I_{1}}-\underbrace{\frac{1}{16}\int\frac{1}{(u-1)(\sqrt{2}u+1)}\;{du}}_{I_{2}}$

    $\displaystyle \displaystyle - ~ \underbrace{\frac{1}{16}\int\frac{1}{(u+1)(\sqrt{2 }u-1)}\;{du}}_{I_{3}}+\underbrace{\frac{1}{16}\int\fr ac{1}{(u+1)(\sqrt{2}u+1)}\;{du}}_{I_{4}}$

    Let $\displaystyle t_{1} = \frac{u-1}{\sqrt{2}u-1}, ~ t_{2} = \frac{u-1}{\sqrt{2}u+1}, ~ t_{3} = \frac{u+1}{\sqrt{2}u-1}, ~ t_{4} = \frac{u+1}{\sqrt{2}u+1}, $ for $\displaystyle I_{1}, I_{2}, I_{3}$
    and $\displaystyle I_{4}$ respectively, then using the quotient rule on each of these, we have:

    $\displaystyle \displaystyle \dfrac{dt_{1}}{du} = \frac{1-\sqrt{2}}{(\sqrt{2}u-1)^2} \Rightarrow \frac{(\sqrt{2}u-1)^2}{1-\sqrt{2}}\;{dt_{1}} = du, ~ \dfrac{dt_{2}}{du} = -\frac{1+\sqrt{2}}{(\sqrt{2}u+1)^2} \Rightarrow -\frac{(\sqrt{2}u+1)^2}{1+\sqrt{2}}\;{dt_{2}} = du$

    $\displaystyle \displaystyle \dfrac{dt_{3}}{du} = \frac{1+\sqrt{2}}{(\sqrt{2}u-1)^2} \Rightarrow \frac{(\sqrt{2}u-1)^2}{1+\sqrt{2}}\;{dt_{3}} = du, ~ \dfrac{dt_{4}}{du} = \frac{\sqrt{2}-1}{(\sqrt{2}u+1)^2} \Rightarrow \frac{(\sqrt{2}u-1)^2}{\sqrt{2}-1}\;{dt_{4}} = du$

    Taking these back to our original integrals, and doing one-by-one we have:

    $\displaystyle \displaystyle I_{1} = \frac{1}{16}\int\frac{1}{(u-1)(\sqrt{2}u-1)}\;{du} = \frac{1}{16(1-\sqrt{2})}\int\frac{(\sqrt{2}u-1)^2}{(u-1)(\sqrt{2}u-1)}\;{dt_{1}} $

    $\displaystyle \displaystyle = \frac{1}{16(1-\sqrt{2})}\int\frac{(\sqrt{2}u-1)}{(u-1)}\;{dt_{1}} = \frac{1}{16(1-\sqrt{2})}\int\frac{1}{t_{1}}\;{dt_{1}} $$$\displaystyle

    $\displaystyle \displaystyle = \frac{1}{16(1-\sqrt{2})}\ln{t_{1}}+k_{1} = \frac{1}{16(1-\sqrt{2})}\ln\left\{\frac{u-1}{\sqrt{2}u-1}\right\}+k_{1}$$$\displaystyle

    $\displaystyle \displaystyle I_{2} = -\frac{1}{16}\int\frac{1}{(u-1)(\sqrt{2}u+1)}\;{du} = \frac{1}{16(1+\sqrt{2})}\int\frac{(\sqrt{2}u+1)^2} {(u-1)(\sqrt{2}u+1)}\;{dt_{2}} $

    $\displaystyle \displaystyle = -\frac{1}{16(1+\sqrt{2})}\int\frac{(\sqrt{2}u+1)}{( u-1)}\;{dt_{2}} = -\frac{1}{16(1+\sqrt{2})}\int\frac{1}{t_{2}}\;{dt_{ 2}}$$$\displaystyle

    $\displaystyle \displaystyle = -\frac{1}{16(1+\sqrt{2})}\ln{t_{2}}+k_{2} = -\frac{1}{16(1+\sqrt{2})}\ln\left\{\frac{u-1}{\sqrt{2}u+1}\right\}+k_{2}$$$\displaystyle

    $\displaystyle \displaystyle I_{3} = \frac{1}{16}\int\frac{1}{(u+1)(\sqrt{2}u-1)}\;{du} = \frac{1}{16(1+\sqrt{2})}\int\frac{(\sqrt{2}u-1)^2}{(u+1)(\sqrt{2}u-1)}\;{dt_{3}} $

    $\displaystyle \displaystyle = \frac{1}{16(1+\sqrt{2})}\int\frac{(\sqrt{2}u-1)}{(u-1)}\;{dt_{3}} = \frac{1}{16(1+\sqrt{2})}\int\frac{1}{t_{3}}\;{dt_{ 3}}$$$\displaystyle

    $\displaystyle \displaystyle = \frac{1}{16(1+\sqrt{2})}\ln{t_{3}}+k_{3} = \frac{1}{16(1+\sqrt{2})}\ln\left\{\frac{u+1}{\sqrt {2}u-1}\right\}+k_{3}$$$\displaystyle

    $\displaystyle \displaystyle I_{4} = \frac{1}{16}\int\frac{1}{(u+1)(\sqrt{2}u+1)}\;{du} = \frac{1}{16(\sqrt{2}-1)}\int\frac{(\sqrt{2}u+1)^2}{(u+1)(\sqrt{2}u+1)}\ ;{dt_{4}} $

    $\displaystyle \displaystyle = \frac{1}{16(\sqrt{2}-1)}\int\frac{(\sqrt{2}u+1)}{(u-1)}\;{dt_{3}} = \frac{1}{16(\sqrt{2}-1)}\int\frac{1}{t_{4}}\;{dt_{4}}$$$\displaystyle

    $\displaystyle \displaystyle = \frac{1}{16(\sqrt{2}-1)}\ln{t_{4}}+k_{4} = \frac{1}{16(\sqrt{2}-1)}\ln\left\{\frac{u+1}{\sqrt{2}u+1}\right\}+k_{4}$$$\displaystyle

    Since $\displaystyle \displaystyle I = I_{1}-I_{2}-I_{3}+I_{4}$$$\displaystyle , we have:

    [Math]\displaystyle I = \frac{1}{16(1-\sqrt{2})}\ln\left\{\frac{u-1}{\sqrt{2}u-1}\right\}+\frac{1}{16(1+\sqrt{2})}\ln\left\{\frac {u-1}{\sqrt{2}u+1}\right\}$

    $\displaystyle \displaystyle -\frac{1}{16(1+\sqrt{2})}\ln\left\{\frac{u+1}{\sqrt {2}u-1}\right\}+\frac{1}{16(\sqrt{2}-1)}\ln\left\{\frac{u+1}{\sqrt{2}u+1}\right\}+k$

    Thus our original integral evaluates to:

    $\displaystyle \displaystyle I = \frac{1}{16(1-\sqrt{2})}\ln\left\{\frac{\sin{x}-1}{\sqrt{2}\sin{x}-1}\right\}+\frac{1}{16(1+\sqrt{2})}\ln\left\{\frac {\sin{x}-1}{\sqrt{2}\sin{x}+1}\right\}$

    $\displaystyle \displaystyle -\frac{1}{16(1+\sqrt{2})}\ln\left\{\frac{\sin{x}+1} {\sqrt{2}\sin{x}-1}\right\}+\frac{1}{16(\sqrt{2}-1)}\ln\left\{\frac{\sin{x}+1}{\sqrt{2}\sin{x}+1}\r ight\}+k$
    Last edited by TheCoffeeMachine; Nov 13th 2010 at 08:43 AM.
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