Mathematics 132 – Calculus for Physical and Life Science 2 Solutions for Discussion 3 – Choosing an Integration Method February 21, 2005

A) This can be done either by trigonometric substitution, or integration by parts. Using the substitution t = 4 sin(θ) (so dt = 4 cos(θ) dθ),

∫ \frac{t^{3}}{\sqrt{16 − t^{2}}} dt = ∫ \frac{(4 sin(θ))^{3} · 4 cos(θ) dθ}{\sqrt{16(1 − (sin(θ))^{2}}}

= ∫ \frac{64 sin^{3}(θ)4 cos(θ) dθ}{4 cos(θ)}

= 64 ∫ sin^{3}(θ) dθ

= 64 (\frac{−1}{3} sin^{2}(θ) cos(θ) − \frac{2}{3} cos(θ)) + C (using #17 in table)

= 64 (\frac{−1}{3} (\frac{t}{4})^{2} · \frac{\sqrt{16 − t^{2}}}{4} − \frac{2}{3} \frac{\sqrt{16 − t^{2}}}{4} ) + C

= \frac{−t^{2}}{3} \sqrt{16 − t^{2}} − \frac{32}{3}\sqrt{16 − t^{2}} + C

(Note: This one can also be done by parts with u = t^{2}!)

B) Rewrite the function algebraically first, splitting into a sum of two terms and cancelling powers of x:

∫ \frac{x + x^{5/2}}{x^{2}} dx = ∫ \frac{1}{x} + x^{1/2} dx = ln |x| + \frac{2}{3}x^{3/2} + C.

C) Note that \frac{d}{dx} \sqrt{x^{2} + 3} = \frac{x}{\sqrt{x^{2}+3}}. So the form of this integral is ∫ sin(u) du if we make the substitution u = \sqrt{x^{2} + 3}. We have

∫ sin(\sqrt{x^{2} + 3})\frac{x}{\sqrt{x^{2} + 3}} dx = ∫ sin(u) du = − cos(u) + C = − cos(\sqrt{x^{2} + 3}) + C.

D) Start by applying integration by parts, with u = arcsin(v) and dw = v^{2} dv. Then

∫ v^{2} arcsin(v) dv = \frac{v^{3}}{3} arcsin(v) − \frac{1}{3} ∫ \frac{v^{3}}{\sqrt{1 − v^{2}}} dv.

The remaining integral now has almost the same form as the integral in question A (except the quadratic inside the square root is 1 − v^{2}, not 16 − v^{2}. So we can finish with a trigonometric substitution v = sin(θ), dv = cos(θ) dθ. We get

= \frac{v^{3}}{3} arcsin(v) − \frac{1}{3} ∫ sin^{3}(θ) dθ

= \frac{v^{3}}{3} arcsin(v) − \frac{1}{3} (\frac{−1}{3} sin^{2}(θ) cos(θ) − \frac{2}{3} cos(θ)) + C

= \frac{v^{3}}{3} arcsin(v) + \frac{1}{9}v^{2}\sqrt{1 − v^{2}} + \frac{2}{9}\sqrt{1 − v^{2}} + C

E) Note that \frac{d}{dx} sin(x^{2}) = cos(x^{2}) · 2x. Hence after the substitution u = sin(x^{2}), this integral becomes

∫_{u=0}^{u=1} u^{5} du = \frac{u^{6}}{12} |_{0}^{1} = \frac{1}{12}.

Note: It is not correct to write the limits of integration on the substituted integral as the original 0 and \sqrt{π/2}. Those are x-values, not u-values. The corresponding values of u are found with the substitution equation:

x = 0 ↔ u = sin(0^{2}) = 0
x = \sqrt{π/2} ↔ u = sin((\sqrt{π/2})^{2}) = sin(π/2) = 1

F) Complete the square, substitute w = x − 2, and use the table of integrals (# 31 and # 29):

\begin{aligned}∫ \sqrt{x^{2} − 4x} dx & = ∫ \sqrt{(x − 2)^{2} − 4} dx \\ & = \frac{x − 2}{2}\sqrt{x^{2} − 4x} − 2 ln |x − 2 + \sqrt{x^{2} − 4x}| + C \end{aligned}

G) This one is done by the method of partial fractions. The denominator factors as u(u^{2} + 8u + 17). So we have

\frac{u + 1}{u(u^{2} + 8u + 17)} = \frac{A}{u} + \frac{Bu + C}{u^{2} + 8u + 17}

For some A, B, C. To find them, we clear denominators to see:

u + 1 = A(u^{2} + 8u + 17) + (Bu + C)u = (A + B)u^{2} + (8A + C)u + 17A.

Then equating coefficients, A + B = 0, 8A + C = 1, 17A = 1. Therefore,

A = \frac{1}{17},  B = \frac{−1}{17},  C = \frac{9}{17}

So,

(1)  ∫ \frac{u + 1}{u(u^{2} + 8u + 17)} du = \frac{1}{17} ∫ \frac{du}{u} + \frac{1}{17} ∫ \frac{−u + 9}{u^{2} + 8u + 17} du

The first term here is easy; we just get \frac{1}{17} ln |u|. For the other term, we complete the square downstairs: u^{2} + 8u + 17 = (u + 4)^{2} + 1, and substitute w = u + 4 (so u = w − 4, and dw = du):

\frac{1}{17} ∫ \frac{−u + 9}{u^{2} + 8u + 17} du = \frac{1}{17} ∫ \frac{−w + 13}{w^{2} + 1} dw

At this point, we can use the table, or just split up the integral into two parts:

\frac{1}{17} ∫ \frac{−w + 13}{w^{2} + 1} dw = \frac{1}{17} ∫ \frac{−w}{w^{2} + 1} dw + \frac{13}{17} ∫ \frac{dw}{w^{2} + 1}
= −\frac{1}{34} ln |w^{2} + 1| + \frac{13}{17} arctan(w) + C

Substituting w = x + 4 back in, and adding the first term above, we get the final answer:

∫ \frac{u + 1}{u(u^{2} + 8u + 17)} du = \frac{1}{17} ln |u| − \frac{1}{34} ln |x^{2} + 8x + 17| + \frac{13}{17} arctan(x + 4) + C