Solutions to Problem Set 5

Math 211-03

9-12-2017

[Trig substitution]

1. Compute $\displaystyle \int
   \dfrac{x^2}{\sqrt{x^2 - 25}}\,dx$ .

Note:

$$\int (\sec \theta)^3\,d\theta = \dfrac{1}{2} \sec \theta \tan \theta + \dfrac{1}{2} \sec \theta \tan \theta + c$$

$$\int \dfrac{x^2}{\sqrt{x^2 - 25}}\,dx = \int \dfrac{25 (\sec\theta)^2}{\sqrt{25 (\sec \theta)^2 - 25}} (5 \sec \theta \tan \theta\,d\theta) = 125 \int \dfrac{1}{\sqrt{25 (\tan \theta)^2}} (\sec \theta)^3 \tan \theta\,d\theta =$$

$$\hfil\raise0.5in\hbox{$\left[x = 5 \sec \theta, \quad dx = 5 \sec \theta \tan \theta\,d\theta\right]$} \hskip0.5in \hbox{\epsfysize=1in \epsffile{secant-triangle5.eps}}\hfil$$

$$125 \int \dfrac{1}{5 \tan \theta} (\sec \theta)^3 \tan \theta\,d\theta = 25 \int (\sec \theta)^3\,d\theta = 25 \left(\dfrac{1}{2} \sec\theta \tan \theta + \dfrac{1}{2} \ln |\sec \theta + \tan \theta|\right) + C =$$

$$\dfrac{25}{2} \left(\dfrac{x \sqrt{x^2 - 25}}{25} + \ln \left|\dfrac{x}{5} + \dfrac{\sqrt{x^2 - 25}}{5}\right|\right) + c.\quad\halmos$$


2. Compute $\displaystyle \int
   \dfrac{1}{\sqrt{16 + x^2}}\,dx$ .

$$\int \dfrac{1}{\sqrt{16 + x^2}}\,dx = \int \dfrac{1}{\sqrt{16 + 16 (\tan \theta)^2}} \cdot 4 (\sec \theta)^2\,d\theta = \int \dfrac{1}{\sqrt{16 (\sec \theta)^2}} \cdot 4 (\sec \theta)^2\,d\theta = \int \dfrac{1}{4 \sec \theta} \cdot 4 (\sec \theta)^2\,d\theta =$$

$$\hfil\raise0.5in\hbox{$\left[x = 4 \tan \theta, \quad dx = 4 (\sec \theta)^2\,d\theta\right]$} \hskip0.5in \hbox{\epsfysize=1in \epsffile{tangent-triangle4.eps}}\hfil$$

$$\int \sec \theta\,d\theta = \ln |\sec \theta + \tan \theta| + c = \ln \left|\dfrac{\sqrt{x^2 + 16}}{4} + \dfrac{x}{4}\right| + c.\quad\halmos$$


3. Compute $\displaystyle \int
   \dfrac{1}{(x^2 + 9)^{3/2}}\,dx$ .

$$\int \dfrac{1}{(x^2 + 9)^{3/2}}\,dx = \int \dfrac{1}{[(3 \tan \theta)^2 + 9]^{3/2}} \cdot 3 (\sec \theta)^2\,d\theta = \int \dfrac{1}{(9 (\tan \theta)^2 + 9)^{3/2}} \cdot 3 (\sec \theta)^2\,d\theta =$$

$$\hfil\raise0.5in\hbox{$\left[x = 3 \tan \theta, \quad dx = 3 (\sec \theta)^2\,d\theta\right]$} \hskip0.5in \hbox{\epsfysize=1in \epsffile{tangent-triangle3.eps}}\hfil$$

$$\int \dfrac{1}{[9 (\sec \theta)^2]^{3/2}} \cdot 3 (\sec \theta)^2\,d\theta = \int \dfrac{1}{27 (\sec \theta)^3} \cdot 3 (\sec \theta)^2\,d\theta = \dfrac{1}{9} \int \dfrac{1}{\sec \theta}\,d\theta = \dfrac{1}{9} \int \cos \theta\,d\theta =$$

$$\dfrac{1}{9} \sin \theta + c = \dfrac{1}{9} \dfrac{x}{\sqrt{x^2 + 9}} + c.\quad\halmos$$


[Partial fractions]

4. Compute $\displaystyle \int
   \dfrac{7 x - 29}{x^2 - 4 x - 21}\,dx$ .

Since $x^2 - 4 x - 21 = (x + 3)
   (x - 7)$ , I have

$$\eqalign{ \dfrac{7 x - 29}{(x + 3) (x - 7)} & = \dfrac{a}{x + 3} + \dfrac{b}{x - 7} \cr \noalign{\vskip2pt} 7 x - 29 & = a (x - 7) + b (x + 3) \cr}$$

Set $x = -3$ . This gives

$$-50 = a \cdot (-10), \quad\hbox{so}\quad a = 5.$$

Set $x = 7$ . This gives

$$20 = b \cdot 10, \quad\hbox{so}\quad b = 2.$$

Therefore,

$$\int \dfrac{7 x - 29}{(x + 3) (x - 7)}\,dx = \int \left(\dfrac{5}{x + 3} + \dfrac{2}{x - 7}\right)\,dx = 5 \ln |x + 3| + 2 \ln |x - 7| + C.\quad\halmos$$


5. Compute $\displaystyle \int
   \dfrac{8 x + 14}{x^2 + 2 x - 8}\,dx$ .

Since $x^2 + 2 x - 8 = (x + 4) (x
   - 2)$ , I have

$$\eqalign{ \dfrac{8 x + 14}{(x + 4) (x - 2)} & = \dfrac{a}{x + 4} + \dfrac{b}{x - 2} \cr \noalign{\vskip2pt} 8 x + 14 & = a (x - 2) + b (x + 4) \cr}$$

Set $x = -4$ . This gives

$$-18 = a \cdot (-6), \quad\hbox{so}\quad a = 3.$$

Set $x = 2$ . This gives

$$30 = b \cdot 6, \quad\hbox{so}\quad b = 5.$$

Therefore,

$$\int \dfrac{8 x + 14}{(x + 4) (x - 2)}\,dx = \int \left(\dfrac{3}{x + 4} + \dfrac{5}{x - 2}\right)\,dx = 3 \ln |x + 4| + 5 \ln |x - 2| + C.\quad\halmos$$


6. Compute $\displaystyle \int
   \dfrac{2 x - 6}{(x - 1)^2}\,dx$ .

Set up the partial fractions decomposition:

$$\eqalign{ \dfrac{2 x - 6}{(x - 1)^2} & = \dfrac{A}{x - 1} + \dfrac{B}{(x - 1)^2} \cr 2 x - 6 & = A (x - 1) + B \cr}$$

Let $x = 1$ . This gives

$$2 \cdot 1 - 6 = B, \quad\hbox{so}\quad B = -4.$$

Therefore,

$$2 x - 6 = A (x - 1) - 4.$$

Let $x = 0$ . This gives

$$-6 = A \cdot (-1) - 4, \quad\hbox{so}\quad A = 2.$$

Thus,

$$\int \dfrac{2 x - 6}{(x - 1)^2}\,dx = \int \left(\dfrac{2}{x - 1} + \dfrac{-4}{(x - 1)^2}\right)\,dx =$$

$$2 \ln |x - 1| + \dfrac{4}{x - 1} + C.\quad\halmos$$


As he thinketh in his heart, so is he. - Proverbs 23:7


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