Problems in calculus and trigonometry

Solution Preview

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)].

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(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)]

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(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) ...