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    Problems in calculus and trigonometry

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    Do the following problems:

    (1) Find the value of twice the integral of the function u^(-1) + 3*[u^(-2)] over the interval [-3, -1].

    (2) Show that cot(pi/3) = 1/[sqrt(3)], where "sqrt" stands for "square root."

    (3) Obtain the Maclaurin series for the function f(x) = sin x.

    (4) Obtain the Maclaurin series for the function f(x) = sin(2*x).

    (5) Obtain the Maclaurin series for the function f(x) = (x^2)*[sin(x^2)].

    (6) Obtain the Maclaurin series for the function f(x) = (x^4) - {(x^2)*[sin(x^2)]}.

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    https://brainmass.com/math/trigonometry/problems-calculus-trigonometry-79853

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    A detailed solution is given for each problem.

    (1) Recall that the antiderivative of u^(-1) is ln(|u|).

    Also, using the Power Law, we find that the antiderivative of 3*u^(-2) is

    3*[u^(-2 + 1)]/(-2 + 1) = 3*[u^(-1)]/(-1) = -3*[u^(-1)].

    Thus the antiderivative of u^(-1) + 3*[u^(-2)] is ln(|u|) - 3*[u^(-1)].

    Evaluating this expression at the limits of integration (upper limit u = -1, lower limit u = -3), we get

    {ln(|-1|) - 3*[(-1)^(-1)]} - {ln(|(-3)| - 3*[(-3)^(-1)]}

    = {ln(1) - 3*[1/(-1)]} - {ln(3) - 3*[1/(-3)]}

    = {0 - [3/(-1)]} - {ln(3) - [3/(-3)]}

    = - [(-3)] - [ln(3) - (-1)]

    = 3 - [ln(3) +1]

    = 3 - ln(3) - 1

    = 2 - ln(3)

    Thus twice the value of this integral is 4 - 2*[ln(3)].

    --------------------------------------------------------------

    (2) To show that cot(pi/3) = 1/[sqrt(3)], use the fact that cot(pi/3) = [cos(pi/3)]/[sin(pi/3)].

    Recall that pi/3 radians is equivalent to 180/3 = 60 degrees, and that cos(60) = 1/2 and sin(60) = [sqrt(3)]/2,

    where "sqrt(3)" stands for "square root of 3".

    Thus

    cot(60) = [cos(60)]/[sin(60] = (1/2)/{[sqrt(3)]/2}

    Inverting the divisor ([sqrt(3)]/2) and multiplying, we find that

    cot(60) = (1/2)*{2/[sqrt(3)]} = 1/[sqrt(3)]

    --------------------------------------------------------------

    (3) Let f(x) = sin x.

    To find the Maclaurin series for f, we take successive derivatives of f, and evaluate those derivatives at x = 0:

    f(x) = sin x; f(0) = sin 0 = 0

    f'(x) = cos x; f'(0) = cos 0 = 1

    f"(x) = -(sin x); f"(0) = -(sin 0) = -(0) = 0

    f'''(x) ...

    Solution Summary

    A complete, detailed solution is given for each problem: For (1), the integral is evaluated; for (2), the value of cot(pi/3) is determined, and found to be 1/[sqrt(3)]; for (3) through (6), the Maclaurin series of the given function is found.

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