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calculus

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Integral of ln(sinx+cosx) with respect to x from -pi/4 to pi/4

  • calculus - ,

    sin(x) + cos(x) = sqrt(2) sin(x + pi/4)

    So, the integral can be written as:

    pi/4 ln(2) + Integral of Log[sin(x)]dx from 0 to pi/2

    Let's call the integral in here I:

    I = Integral of Log[sin(x)]dx from 0 to pi/2

    Then note that because sin is symmetric w.r.t. reflection about x = pi/2 integrating to pi will yield twice the value:

    2 I = Integral of Log[sin(x)]dx from 0 to pi

    Then substitute in this integral
    x = 2 y:

    2 I = 2 Integral of Log[sin(2y)]dy from 0 to pi/2 ----->


    I = Integral of
    (Log(2) + Log[cos(y)] +Log[sin(y)] )dy from 0 to pi/2 =

    pi/2 Log(2) + 2 I

    because the integral of log[cos(y)] is the same as the integral of
    log[sin(y)], as cos(y) is sin(pi/2-y).

    Solving for I gives:

    I = -pi/2 log(2)

    The original integral is thus
    - pi/4 log(2)

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