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Calculus 2

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integral from 0 to infinity of 1/(pi^2+x^2)^(3/2)dx

  • Calculus 2 -

    if x = pi*tan(z)
    pi^2 + x^2 = pi^2 + pi^2 tan^2(z) = pi^2 sec^2(z)
    dx = pi*sec^2(z)

    1/(pi^3 sec^3(z)) * pi*sec^2(z) dz
    = 1/pi^2 cos(z) dz

    the integral is thus 1/pi^2 sin(z)

    x/pi * tan(z), so
    sin(z) = x/sqrt(pi^2 + x^2)

    the integral is thus x/[pi^2 sqrt(pi^2 + x^2)]

    from 0 to infinity is thus 1/pi^2

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