Mathematics 132 – Calculus for Physical and Life Sciences 2 Makeup Quiz Solutions – February 29, 2008

Some possibly useful formulas:

sin^{2} θ + cos^{2} θ = 1,  tan^{2} θ + 1 = sec^{2} θ,  sin(2θ) = 2 sin θ cos θ.

(5) A) Integrate:

∫ \frac{x^{3}}{\sqrt{9 − x^{2}}} dx

Because of the 9−x^{2} under the radical sign, we use the sin substitution, letting x = 3 sin θ. Then dx = 3 cos θ dθ and we proceed as follows:

∫ \frac{x^{3}}{\sqrt{9 − x^{2}}} dx = ∫ \frac{(3 sin θ)^{3}}{\sqrt{9 − (3 sin θ)^{2}}} 3 cos θ dθ

= ∫ \frac{27 sin^{3} θ · 3 cos θ}{\sqrt{9(1 − sin^{2} θ)}} dθ

= ∫ \frac{27 sin^{3} θ · 3 cos θ}{\sqrt{9 cos^{2} θ}} dθ (by the basic identity)

= ∫ \frac{27 sin^{3} θ · 3 cos θ}{3 cos θ} dθ

= 27 ∫ sin^{3} θ dθ (after cancelling)

= 27 ∫ sin^{2} θ · sin θ dθ (odd power strategy)

= 27 ∫ (1 − cos^{2} θ) sin θ dθ

= 27 (∫ sin θ dθ − ∫ cos^{2} θ sin θ dθ) (multiply out and split)

= 27 (− cos θ + \frac{1}{3} cos^{3} θ) + C

The last line comes from applying the u-substitution u = cos θ to the second integral (note that du = − sin θ dθ). Now we convert back to functions of x from the substitution x = 3 sin θ. The reference triangle has opposite side x, hypotenuse 3 and adjacent side \sqrt{9 − x^{2}}. Hence cos θ = \sqrt{9 − x^{2}}/3, and the integral equals

= −9\sqrt{9 − x^{2}} + \frac{1}{3}(9 − x^{2})^{3/2} + C.

(5) B) Integrate:

∫ \frac{x^{4} + 2x + 1}{x^{3} + x} dx

Since the degree of the top is larger, we have to divide the denominator into the numerator. The quotient is x and the remainder is x^{4} + 2x + 1 − x(x^{3} + x) = −x^{2} + 2x + 1. Therefore,

\frac{x^{4} + 2x + 1}{x^{3} + x} = x + \frac{−x^{2} + 2x + 1}{x^{3} + x} = x + \frac{−x^{2} + 2x + 1}{x(x^{2} + 1)}.

We split the second term up into partial fractions like this:

\frac{−x^{2} + 2x + 1}{x(x^{2} + 1)} = \frac{A}{x} + \frac{Bx + C}{x^{2} + 1} , (so after clearing denoms)
−x^{2} + 2x + 1 = A(x^{2} + 1) + x(Bx + C)
−x^{2} + 2x + 1 = (A + B)x^{2} + Cx + A

By comparing coefficients A = 1, C = 2 and A + B = −1. Hence B = −2. This gives

x + \frac{−x^{2} + 2x + 1}{x(x^{2} + 1)} = x + \frac{1}{x} + \frac{−2x + 2}{x^{2} + 1} (so)
∫ x + \frac{−x^{2} + 2x + 1}{x(x^{2} + 1)} dx = ∫ x + \frac{1}{x} + \frac{−2x + 2}{x^{2} + 1} dx

= \frac{x^{2}}{2} + ln |x| + ∫ \frac{−2x}{x^{2} + 1} dx + ∫ \frac{2}{x^{2} + 1} dx

= \frac{x^{2}}{2} + ln |x| − ln(x^{2} + 1) + 2 tan^{−1}(x) + C

The third term here comes from the integral ∫ \frac{−2x}{x^{2}+1} dx, which is − ∫ \frac{du}{u} for u = x^{2} + 1. The last term is the basic inverse tangent integral.