? asked in Science & MathematicsMathematics · 1 decade ago

How do you integrate cos(x)/(4-sin^2(x)) from 0 to 1?

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  • Anonymous
    1 decade ago
    Favorite Answer

    sin(x) = 2 tanh(u)

    cos(x) dx = 2sech^2(u) du

    u = arctanh(sin(x))/2)

    ∫cos(x) dx /(4-sin^2(x))

    = ∫2sech^2(u) du / (4 - 4 tanh^2(u))

    = ∫2sech^2(u) du / [(4cosh^2 - 4 sinh^2(u)) / cosh^2(u) )

    = ∫2sech^2(u) du / [(4) / cosh^2(u) )

    = ∫2sech^2(u) du / [(4) sech^2(u) )

    = (1/2) ∫ du = u / 4

    = (1/2) arctanh(sin(x))/2)

    developing an expression for arctanh

    z = arctanh(w)

    w = tanh(z) = (e^z - e ^ -z) / (e^z + e^-z) = (e^2z - 1)/ (e^2z + 1)

    w e^2z + w = e^2z - 1

    e^2z(w - 1) = -1 - w

    e^2z = (1 + w)/(1-w)

    z = (1/2) log((1 + w)/(1-w))

    arctanh(w) = (1/2) log((1 + w)/(1-w))

    Final answer for integral

    ∫cos(x) dx /(4-sin^2(x)) = (1/2) arctanh(sin(x))/2)

    = (1/4) log((1 + sin(x)/2)/(1-sin(x)/2)

    = (1/4) log((2 + sin(x))/(2 - sin(x)) + C

    Evaluate from 0 to 1

    (1/4) log((2 + sin(1))/(2 - sin(1)) - (1/2) log((2 + sin(0))/(2 - sin(0))

    = .22429 - 0 = .22429

    = .448585

    Source(s): if you've got an arctan button on your calculator, you can change limits then substitute
  • 1 decade ago

    =1/4ln(sinx+2)/(sinx-2) from(0,1)=

    =1/4[ln(sin1+2)/(sin1-2)-(ln(sin0+2)/(sin0-2)]=

    =1/4[ln(2.017/-1.98)-ln(2/-2)]=infiniti

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