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Pe: 2017-12-03T15:37:26+02:00 In: In:

Integrala

\int_{0}^{\pi }sinx\cdot cosx\cdot cos2x\cdot cos4x\cdot ...\cdot cos2^{n-1}x

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3 raspunsuri

  1. pp

    P=sinx\cdot cosx\cdot cos2^{1}x\cdot cos2^{2}x\cdot .......cos2^{n-1}x/\cdot 2^{n}
    ->>P\cdot 2^{n}=2sinx\cdot cosx\cdot cos2^{1}x\cdot cos2^{2}x\cdot ....... \cdot cos2^{n-1}x\cdot 2^{n-1}
    P\cdot 2^{n}=sin2^{1}x\cdot cos2^{1}x\cdot cos2^{2}x\cdot ....... \cdot cos2^{n-1}x\cdot 2^{n-1}
    P\cdot 2^{n}=2\cdot sin2^{1}x\cdot cos2^{1}x\cdot cos2^{2}x\cdot ....... \cdot cos2^{n-1}x\cdot 2^{n-2}
    P\cdot 2^{n}=sin2^{2}x\cdot cos2^{2}x\cdot ....... \cdot cos2^{n-1}x\cdot 2^{n-2}

    …………………………………………………………………………………………………………………….
    P\cdot 2^{n}=2\cdot sin2^{n-1}\cdot cos2^{n-1}x=sin2^{n}x
    ->P=\frac{sin2^{n}x}{2^{n}}
    Apoi aplici integrala din P.

  2. DD
    profesor

    Exprsia de integrat se poate scrie si;
    Sinx.cosx.cos2x.cos4x……cos2^(n-1)x=(1/2^n)sin2^nx.
    (1/2^n) sin(2^n).x=x–2^(2n). x^3/3!+2^(4n). x^5/5!–2^(6n). x^7/7!+…….
    ∫_0^π▒(1/2^n) sin(2^n) x.dx.=π^2/2–2^(2n). π^4/4!+2^(4n). π^6/6!–2^(6n). π^8/8!+……

  4. DD
    profesor

    Revenire si scuze.
    ∫_0^π▒(1/2^n ) *sin(2^n )x*dx=(1/2^2n )*cos(2^n )x|_0^π=(1/2^2n )*(1-1)=0

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