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Math Help - 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{\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{\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{\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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  5. #5
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    Quote Originally Posted by tonio View Post
    \int\frac{dx}{4\cos x(2\cos^2x-1)}\,dx}
    \displaystyle\frac{1}{\cos x(2{{\cos }^{2}}x-1)}=\frac{\cos x}{(1-{{\sin }^{2}}x)(1-2{{\sin }^{2}}x)}, substitute t=\sin t and get

    \begin{aligned}<br />
   \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} \\ <br />
 & =2\int{\frac{dt}{1-2{{t}^{2}}}}-\int{\frac{dt}{1-{{t}^{2}}}\,dt}, \\ <br />
\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 u = \sin{x} gives  \displaystyle \int\frac{1}{8 u^4-12 u^2+4}\;{du} , then simply write:

    [LaTeX ERROR: Convert failed] " alt="\displaystyle 8u^4-12u^2+4 = 8t^2-12t+4 = 4(2t^2-3t+1)[/Math]

    [LaTeX ERROR: Convert failed] " />[LaTeX ERROR: Convert failed] <br /> <br />
[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}

    [LaTeX ERROR: Code too long, max. 1000 characters] [LaTeX ERROR: Convert failed] " alt="

    [LaTeX ERROR: Convert failed] " />[LaTeX ERROR: Convert failed] [LaTeX ERROR: Convert failed] " alt="

    [LaTeX ERROR: Convert failed] " />[LaTeX ERROR: Convert failed] [LaTeX ERROR: Convert failed] " alt="

    [LaTeX ERROR: Convert failed] " />[LaTeX ERROR: Convert failed] [LaTeX ERROR: Convert failed] " alt="

    [LaTeX ERROR: Convert failed] " />[LaTeX ERROR: Convert failed] , we have: <br /> <br />
[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 -\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 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 -\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; November 13th 2010 at 09:43 AM.
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