]> Haese & Harris - Chapter 29 Notes

29: Further Integration; Differential Equations

A: Integrals resulting in Arcfunctions

We all know that $\frac{d}{d x} [{tan}^{- 1} (x)] = \frac{1}{x^{2} + 1}$ ; thus, we know that $\int \frac{1}{x^{2} + 1} d x = {tan}^{- 1} (x) + C$ . The trick (problem) is that some integrals have the pattern for the arcfunction hidden, and must be worked on in order to reveal the pattern we need.

I'm not going to follow the two patterns in the text—they are too restrictive; you need to be more flexible!

The Patterns

Important basic integrals!

$\int \frac{1}{x^{2} + 1} d x = {tan}^{- 1} (x) + C$

$\int \frac{1}{\sqrt{1 - x^{2}}} d x = {sin}^{- 1} (x) + C$

$\int \frac{1}{| x | \sqrt{x^{2} - 1}} d x = {sec}^{- 1} (x) + C$

Examples

[1.] Find $\int \frac{1}{x^{2} + 9} d x$ .

This is in the pattern for arctangent, but the 9 is a problem—it needs to be a one! The solution begins with factoring nine out of the denominator: $\int \frac{1}{9 (\frac{x^{2}}{9} + 1)} d x$ . Let's clean that up a bit: $\frac{1}{9} \int \frac{1}{{(\frac{x}{3})}^{2} + 1} d x$ . Now, perhaps, you can see what substitution is required to finish? Let $u = \frac{x}{3}$ so that $d u = \frac{1}{3} d x \Rightarrow 3 d u = d x$ . The integral becomes $\frac{1}{9} \cdot 3 \int \frac{1}{u^{2} + 1} d u = \frac{1}{3} {tan}^{- 1} (u) + C = \frac{1}{3} {tan}^{- 1} (\frac{x}{3}) + C$ .

[2.] Find $\int \frac{1}{x \sqrt{x^{2} - 4}} d x$ .

This is in the arcsecant pattern…except for the four. What will you do?

$\int \frac{1}{x \sqrt{4 (\frac{x^{2}}{4} - 1)}} d x = \int \frac{1}{2 x \sqrt{{(\frac{x}{2})}^{2} - 1}} d x$ . This is ripe for substitution! Let $u = \frac{x}{2}$ so that $d u = \frac{1}{2} d x \Rightarrow 2 d u = d x$ . Be careful—since $u = \frac{x}{2}$ , $2 u = x$ . Substitute! $2 \int \frac{1}{2 (2 u) \sqrt{u^{2} - 1}} d u = \frac{1}{2} \int \frac{1}{u \sqrt{u^{2} - 1}} d u$ , which makes $\frac{1}{2} {sec}^{- 1} (u) + C = \frac{1}{2} {sec}^{- 1} (\frac{x}{2}) + C$ .

[3.] Find $\int \frac{1}{\sqrt{4 x^{2} - 9}} d x$ .

Now there are two problems…focus on the goal! Factor out a 9 from the denominator. That makes $\frac{1}{3} \int \frac{1}{\sqrt{\frac{4}{9} x^{2} - 1}} d x$ , which is the same as $\frac{1}{3} \int \frac{1}{\sqrt{{(\frac{2 x}{3})}^{2} - 1}} d x$ . Let $u = \frac{2 x}{3}$ so that $d u = \frac{2}{3} d x \Rightarrow \frac{3}{2} d u = d x$ . The integral becomes $\frac{1}{2} \int \frac{1}{\sqrt{u^{2} - 1}} d u$ , which evaluates to $\frac{1}{2} {sin}^{- 1} (u) + C = \frac{1}{2} {sin}^{- 1} (\frac{2 x}{3}) + C$ .

B: Further Integration by Substitution

There is a difference between knowing the path and walking the path…

There are far too may ways to do u-substitution to count. Your best bet is to jump in there and start working some. Here are a few examples to spark some ideas…be sure to note the excellent advice box on page 713 of the text.

Examples

[4.] Find $\int \frac{ln (x^{2})}{x} d x$ .

Letting $u = x^{2}$ is tempting…but wrong. Let $u = ln (x^{2})$ , so that $d u = \frac{1}{x^{2}} \cdot 2 x \cdot d x = \frac{2}{x} d x$ . The integral becomes $\frac{1}{2} \int u \cdot d u = \frac{1}{4} u^{2} + C = \frac{1}{4} ln (x^{2}) + C$ .

[5.] Find $\int \frac{1}{x^{2} \sqrt{x^{2} - 9}} d x$ .

I'll try a trigonometric substitution—let $x = 3 sec (θ)$ , so that $d x = 3 sec (θ) tan (θ) d θ$ . The integral becomes $\int \frac{3 sec (θ) tan (θ)}{9 {sec}^{2} (θ) \sqrt{9 {sec}^{2} (θ) - 9}} d θ$ . Cleaning that up results in $\frac{1}{9} \int \frac{tan (θ)}{sec (θ) \sqrt{{sec}^{2} (θ) - 1}} d θ$ . The radicand is an identity, making $\frac{1}{9} \int \frac{tan (θ)}{sec (θ) \sqrt{{tan}^{2} (θ)}} d θ$ . That leads to $\frac{1}{9} \int \frac{1}{sec (θ)} d θ = \frac{1}{9} \int cos (θ) d θ = \frac{1}{9} sin (θ) + C$ . I should put this back in terms of the original variable: $\frac{1}{9} sin ({sec}^{- 1} (\frac{x}{3})) + C$ .

That combination of trig functions is something that you did (should have done) before! I'm going to draw a picture of the angle θ:

Triangle

That makes the other side $\sqrt{x^{2} - 9}$ , so $sin (θ) = \frac{\sqrt{x^{2} - 9}}{x}$ …thus, the integral answer reduces to $\frac{\sqrt{x^{2} - 9}}{x} + C$ .

[6.] Find $\int \sqrt{16 - 4 x^{2}} d x$ .

This will be a little unusual—let $x = 2 sin (θ)$ , so that $d x = 2 cos (θ) d θ$ . That makes the integral $\int \sqrt{16 - 16 {sin}^{2} (θ)} \cdot 2 cos (θ) d θ$ . Factor in the radicand: $8 \int \sqrt{1 - {sin}^{2} (θ)} \cdot cos (θ) d θ$ . The radicand $8 \int \sqrt{{cos}^{2} (θ)} \cdot cos (θ) d θ = 8 \int {cos}^{2} (θ) d θ$ . This will require a power-reducing identity—specifically ${cos}^{2} (θ) = \frac{1}{2} + \frac{1}{2} cos (2 θ)$ . The integral now becomes $8 \int (\frac{1}{2} + \frac{1}{2} cos (2 θ)) d θ = 4 \int (1 + cos (2 θ)) d θ$ . This can be done: $4 θ + \frac{1}{2} sin (2 θ) + C$ . Now to put it back into terms of the original variable—since $x = 2 sin (θ)$ , $θ = {sin}^{- 1} (\frac{x}{2})$ , and the answer is $4 {sin}^{- 1} (\frac{x}{2}) + \frac{1}{2} sin ({sin}^{- 1} (\frac{x}{2})) = 4 {sin}^{- 1} (\frac{x}{2}) + \frac{1}{2} \cdot \frac{x}{2} = 4 {sin}^{- 1} (\frac{x}{2}) + \frac{x}{4}$ .

C: Integration by Parts

This is the product rule, in reverse! It is useful for untangling certain integrals.

Since $\frac{d}{d x} [f \cdot g] = f^{'} g + f g^{'}$ , $f \cdot g = \int g d f + \int f d g$ . This is usually written in a equivalent for $\int u d v = u \cdot v - \int v d u$ .

At first glance, you might think that this hasn't done anything—one integral has been substituted for another. That is correct, but since the variables have changed, new integration opportunities may open up as a result…

The real trick is in deciding which part will be u, and which part will be dv. Much like u-substitution, you'll just have to get a feel for it!

Examples

[7.] Find $\int x e^{x} d x$ .

I'll let $u = x$ and $d v = e^{x} d x$ . That makes $d u = d x$ and $v = e^{x}$ . The integral becomes $x e^{x} - \int e^{x} d x$ , which reduces to $x e^{x} - e^{x} + C$ .

[8.] Find $\int x^{2} ln (x) d x$ .

There is no choice here—I've got to use $u = ln (x)$ and $d v = x^{2} d x$ , so that $d u = \frac{1}{x} d x$ and $v = \frac{1}{3} x^{3}$ . The integral becomes $ln (x) \frac{1}{3} x^{3} - \int \frac{1}{3} x^{3} \cdot \frac{1}{x} d x = \frac{1}{3} x^{3} ln (x) - \frac{1}{3} \int x^{2} d x$ . The answer is $\frac{1}{3} x^{3} ln (x) - \frac{1}{9} x^{3} + C$ .

[9.] Find $\int x^{2} sin (x) d x$ .

This one will be interesting! Let $u = x^{2}$ and $d v = sin (x) d x$ , so that $d u = 2 x d x$ and $v = - cos (x)$ . The integral becomes $- x^{2} cos (x) - \int - 2 x cos (x) d x$ . This looks bad, but don't despair—try again! This time, let $u = x$ and $d v = cos (x) d x$ , so that $d u = d x$ and $v = sin (x) d x$ . The integral was $- x^{2} cos (x) + 2 \int x cos (x) d x$ ; now it becomes $- x^{2} cos (x) + 2 (x sin (x) - \int sin (x) d x)$ . This is workable! The answer is $- x^{2} cos (x) + 2 x sin (x) + 2 cos (x) + C$ .

D: Separable Differential Equations

An equation that includes a derivative is called a differential equation. For example, $\frac{d y}{d x} = x^{2}$ or $\frac{d z}{d t} = t + z$ .

The highest order derivative that appears in the equation is the order of the equation. Both of the preceding examples are first order. We will deal with very simple first and second order differential equations. A simple second order differential equation is $\frac{d^{2} s}{d t^{2}} = - 9.81$ .

In particular, we will deal with differential equations that are separable—the variables can be completely separated to opposite sides of the equation. The equation $\frac{d y}{d x} = x^{2}$ can be written as $d y = x^{2} d x$ , so it is separable. The equation $\frac{d z}{d t} = t + z$ cannot be separated—once you move the differential operator over, there's nothing else you can do!

Solving a differential equation involves taking the integral. The constant of integration need only be placed on one side of the equation. Doing this gives the general solution to the differential equation (the solution of a differential equation is another equation!)

If you are given some information about a specific solution—if you are given some initial conditions—then you can solve for the constant of integration and find a specific solution for the differential equation.

Examples

[10.] Verify that $y = C e^{4 x}$ is a solution to the differential equation $\frac{d y}{d x} = 4 y$ .

First of all, take the derivative of the candidate solution: $\frac{d y}{d x} = 4 C e^{4 x}$ . Now, notice that most of that is just the original equation: $\frac{d y}{d x} = 4 C e^{4 x} = 4 (C e^{4 x}) = 4 y$ . That's it—the given equation is a solution of the differential equation.

[11.] Solve $\frac{d y}{d x} = x y$ .

Separate the variables: $\frac{1}{y} d y = x d x$ . Integrate: $ln | y | = \frac{1}{2} x^{2} + C$ . We should solve this for y: $| y | = e^{\frac{1}{2} x^{2} + C}$ . This can be simplified: $| y | = e^{\frac{1}{2} x^{2}} \cdot e^{C}$ . Note that $e^{C}$ is a constant—thus, we cen simply replace it with the letter C. The added benefit is that this allows the removal of the absolute value! The general solution is $y = C e^{\frac{1}{2} x^{2}}$ .

[12.] Solve $y \frac{d y}{d x} - e^{x} = 0$ if $y (0) = 4$ .

First, the general solution: $y d y = e^{x} d x \Rightarrow \frac{1}{2} y^{2} = e^{x} + C \Rightarrow y = \pm \sqrt{2 e^{x} + C}$ (note that there is no need to write the constant as 2C). Now, plug in the given information and solve for C: $4 = \pm \sqrt{2 e^{0} + C} \Rightarrow 4 = \pm \sqrt{2 + C} \Rightarrow 16 = 2 + C \Rightarrow C = 14$ .

You might think that the solution is $y = \pm \sqrt{2 e^{x} + 14}$ —but it's not. Notice that $4 \neq - \sqrt{2 e^{0} + 14} = - 4$ ; thus, we only need the positive version.

The solution is $y = \sqrt{2 e^{x} + 14}$ .

Page last validated 2010-08-15