Chapter 7 Techniques of Integration and Differential Equations
7.1 Basic Integration Rules
The two principal techniques for integration are substitution and integration by parts.The method of substitution was introduced in Section 5.4; we have used it occasionally in Chapters 5 and 6.
Standard Forms
A short list of known integrals shown below is so useful that we think that every calculus student should memorize it.
Substitution in Indefinite and Definite Integrals
This section is a review of techniques we learnt and used in the previous chapters. We go over all examples on board in class.
Trigonometric Integrals
Manipulating the Integrand
7.2 Integration by Parts
Integration by Parts: Indefinite Integrals
\[ \int u dv= uv-\int v du \]
Integration by Parts: Definite Integrals \[\int_a^b u dv= [uv]\Bigg|_a^b-\int_a^b v du\]
These formulas allow us to shift the problem of integrating \(udv\) to that of integrating \(vdu\). Success depends on the proper choice of \(u\) and \(dv\) which comes with practice.
Go through Example 1-4 on board. There are simple but helpful practice.
Repeated Integration by Parts
Sometimes it is necessary to apply integration by parts several times.
N.B. The fact that the integral we wanted to find reappeared on the right side is what made Example 6 work.
The formula found in Example 7 is called a reduction formula because it replaces an integral containing some power of a function with an integral of the same form having the power reduced.
When \(n\) is a positive integer, we may apply the formula repeatedly until the remaining integral is easy to evaluate. In addition, this iterative procedure can be used to evaluate the definite integral, for example,
7.3 Some Trigonometric Integrals
We consider four commonly encountered types. (In the textbook, there are equivalently five types listed.)
Products of Powers of Sines and Cosines \[\int \sin^mx \cos^nxdx\] where \(m\) and \(n\) are nonnegative integers (positive or zero).
Products of Sines and Cosines \[\int \sin mx\cos nx dx, \int\sin mx\sin nx dx, \int\cos mx \cos nx dx\]
Integrals of Powers of \(\tan x\) or \(\cot x\) \[\int \tan^n xdx, \int \cot^nxdx\]
Products of Powers of \(\tan x\) and \(\sec x\), Powers of \(\cot x\) and \(\cot x\)\[\int\tan^mx\sec^nxdx, \int\cot^mx\csc^nxdx\]
Products of Powers of Sines and Cosines
We begin with integrals of the form \[\int \sin^mx \cos^nxdx\] where \(m\) and \(n\) are nonnegative integers (positive or zero). We can divide the appropriate substitution into three cases according to \(m\) and \(n\) being odd or even.
Example 1, page 397. Case 1. Find \(\int \sin^5xdx\)
Example 3, page 397. Case 1. Find \(\int \sin^3x\cos^{-4}xdx\).
Example of Case 2. Find \(\int\cos^5xdx\).
Example 4, page 398. Case 3. Find \(\int \sin^2x\cos^4xdx\)
Products of Sines and Cosines
The integrals \[\int \sin mx\cos nx dx, \int\sin mx\sin nx dx, \int\cos mx \cos nx dx\]arise in many applications involving periodic functions. We can evaluate these integrals through integration by parts, but two such integrations are required in each case. It is simpler to use the identities
Example 5, using Identity 1. Find \(\int\sin2x\cos3xdx\).
Example 6, using Identity 2.
Integrals of Powers of \(\tan x\) or \(\cot x\)
For the integrals in the tangent case, \[\int \tan^n xdx, \int \cot^nxdx\] factor out \(\tan^2x=\sec^2-1\), and \(\cot^2x=\csc^2x-1\), respectively, for \(n\geq 2\).
Products of Powers of \(\tan x\) and \(\sec x\), Powers of \(\cot x\) and \(\cot x\)
\[\int\tan^mx\sec^nxdx, \int\cot^mx\csc^nxdx\]
When \(n\) is even, \(m\) is any number. Factor out \(\sec^2 x\) and let \(du=\sec^2xdx\), \(u=\tan x\). Using identity \(\sec^2=\tan^2x+1\), find \(\int\tan^{-3/2}x\sec^4xdx\), as shown in example 10, page 401.
When \(m\) is odd, \(n\) is any number. Factor out \(\sec x\tan x\), and let \(du=\sec x\tan xdx\), \(u=\sec x\). Using identity \(\tan^2x=\sec^2-1\), find \(\int \tan^3x\sec^{-1/2}xdx\), as shown in example 11, page 401.
7.4 Rationalizing Substitutions
Radicals in an integrand are often troublesome and we usually try to get rid of them.
Integrands Involving \(\sqrt[n]{ax+b}\)
The substitution \(u=\sqrt[n]{ax+b}\) will eliminate the radical.
Integrands Involving \(\sqrt{a^2-x^2}, \sqrt{a^2+x^2}\) and \(\sqrt{x^2-a^2}\)
N.B. Drawing a graph to help us replace \(t\) and \(\cos t\) in terms of \(x\).
Drawing a graph to help us replace \(\sec t\) and \(\tan t\) in terms of \(x\).
A Supplementary Integral Form
\[ \int \sec xdx=\ln |\sec x+\tan x|+C \]
proof.
\[\begin{aligned} \sec x&=\frac{1}{\cos x}=\frac{1+\sin x}{\cos x(1+\sin x)}\\&=\frac{\sin^2 x+\cos^2x+\sin x}{\cos x(1+\sin x)}\\&=\frac{\sin x(1+\sin x)+\cos^2x}{\cos x(1+\sin x)}\\&=\frac{\sin x}{\cos x}+\frac{\cos x}{(1+\sin x)}\end{aligned}\]
Let \(u=\cos x\) and \(v=1+\sin x\)\[\begin{aligned}\int \sec x dx&=\int \frac{\sin x}{\cos x}dx+\int\frac{\cos x}{(1+\sin x)}dx\\&=-\int \frac{1}{\cos x}d(\cos x)+\int\frac{1}{1+\sin x}d(1+\sin x)\\&=-\ln |u|+\ln |v|+C\\&=\ln\left|\frac{\cos x}{1+\sin x}\right|+C\\&=\ln|\sec x+ \tan x|+C\end{aligned}\]
Completing the Square
When a quadratic expression of the type \(x^2+Bx+C\) appears under a radical, completing the square will prepare it for a trigonometric substitution.
Example 7 (a), page 405. Find \(\int \frac{dx}{\sqrt{x^2+2x+26}}\).
Complete the square, \((x+1)^2+25\), and let \(u=x+1\) and \(du=dx\), then use the trigonometric substitution tricks in previous sub-section to solve \[\int \frac{du}{\sqrt{u^2+25}}\]
7.5 Integration of Rational Functions Using Partial Fractions
A rational function is by definition the quotient of two polynomial functions. Proper rational functions means that the degree of the numerator is less than that of the denominator.
An improper rational function can always be written as a sum of a polynomial function and a proper rational function via long division. For example,
Since polynomials are easy to integrate, the problem of integrating rational functions is really that of integrating proper rational functions.
Partial Fraction Decomposition (Linear Factors)
It is a standard algebraic exercise to add fractions: find a common denominator and add. Now we are interested in the reverse process, that is, decompose a fraction into a sum of simpler fractions. For example, to decompose \((3x-1)/(x^2-x-6)\), we are hoping for a decomposition of the following form: \[\frac{3x-1}{(x+2)(x-3)}=\frac{A}{x+2}+\frac{B}{x-3}\] Therefore,
When we are done with the decomposition, we can find the indefinite integral of the fraction.
Example 4, page 409. Decompose into three terms. Here is a easier way to determine the value of numerators.
N.B. The general rule for decomposing fractions with repeated linear factors in the denominator is this: for each factor \((ax+b)^k\) of the denominator, there are \(k\) terms in the partial fraction decomposition: \[ \frac{A_1}{ax+b}+\frac{A_2}{(ax+b)^2}+\frac{A_3}{(ax+b)^3}+\ldots+\frac{A_k}{(ax+b)^k}\]
Partial Fraction Decomposition (Quadratic Factors)
To determine the constants A, B, C, D and E, we multiply both sides by \((x+3)(x^2+2)^2\) and obtain \[6x^2-15x+22=A(x^2+2)^2+(Bx+C)(x+3)(x^2+2)+(Dx+E)(x+3)\]
Substitution of \(x=-3,x=0, x=1\) and \(x=-1\) yields \[\begin{aligned}121&=121A &(1)\\22&=4+6C+3E&(2)\\13&=9+12(B+C)+4(D+E)&(3)\\43&=9+6(-B+C)+2(-D+E)&(4)\end{aligned}\]
Adding equation (3) by two times equation (4), we find \(C=3\) and \(E=0\), then with eqn. (2) and (3) \(B=-1, D=-5\), . From eqn. (1), \(A=1\).
Therefore,
The Logistic Differential Equation
In Section 4.10, we saw that the assumption that the rate of growth of a population is proportional to its size, that is, \(y^\prime=ky\), leads to exponential growth. When the available resources in the system are unable to sustain the population, more reasonable assumptions are that there is a maximum capacity \(L\) that the system can sustain, and that the rate of growth is proportional to the product of the population size \(y\) and the “available room” \(L-y\). These assumptions lead to the differential equation \[y^\prime=ky(L-y)\]This is called the logistic differential equation.
Now that we have covered the method of partial fractions, we can perform the necessary integration to solve this differential equation.
Example 9, page 411. Solve \(y^\prime=0.0003y(2000-y)\) with the initial condition \(y(0)=800\).
7.7 First-Order Linear Differential Equations
In Section 4.10 we used the method of separation of variables to solve differential equations involving growth and decay. Not all differential equations are separable. For example, in the differential equation \[\frac{dy}{dx}=2x-3y\] The differential equation of the form \[\frac{dy}{dx}+P(x)y=Q(x)\] where \(P(x)\) and \(Q(x)\) are functions of \(x\) only, is called a first-order linear differential equation.
A review of terminologies:
The family of all solutions of a differential equation is called the general solution.
A function that satisfies the differential equation and the initial condition is called a particular solution.
Solving First-Order Linear Equations
Multiply both sides by the integrating factor \[e^{\int P(x)dx}\] the differential equation is then
Example 1, page 425. Solve \[\frac{dy}{dx}+\frac{2}{x}y=\frac{\sin 3x}{x^2}\]
Let’s start with calculating the integrating factor,
\[ e^{\int P(x)dx}=e^{\int \frac{2}{x}dx}=e^{2\ln |x|}=e^{\ln x^2}=x^2 \]
Note here we choose the arbitrary constant from the integration \(e^{\int P(d)dx}\) to be \(0\). That does not affect the answer to the differential equation.
Multiplying both sides of the original equation by the integrating factor, we obtain \[x^2\frac{dy}{dx}+2xy=\sin 3x\]
In example 1, we find a general solution to the given differential equation. Now let’s look at another example that require a particular solution, that is, specify the value of constant \(C\).
Applications
Example 3, page 424.
Let \(y\) measure the quantity of interest that is in the tank at time \(t\). The general principle is \[\frac{dy}{dt}=\text{rate in}-\text{rate out}\]
RL Circuits
The figure above represents an electrical circuit whose total resistance is a constant \(R\) ohms and whose self-inductance, shown as a coil, is \(L\) henries, also a constant. There is a switch whose terminals at \(a\) and \(b\) can be closed to connect a constant electrical source of \(V\) volts or a voltage of \(V(t)\) volts at time \(t\).
Ohm’s Law, \(V = RI\), has to be augmented for such a circuit. The correct equation of Kirchhoff’s Law accounting for both resistance and inductance is \[L \frac{dI}{dt} + RI = V(t),\] where \(I\) is the current measured in amperes and \(t\) is the time in seconds.
This is a linear equation, easily solved by the method of this section. By solving this equation, we can predict how the current will flow after the switch is closed.
Example 4, page 425.
Multiply both sides of the equation by the integrating factor \(e^{\int P(t)dt}=e^{\int 3dt}=e^{3t}\), we obtain \[\int\left(e^{3t}\frac{dI}{dt}+3e^{3t}I\right)dt=\int 6e^{3t}dt\\e^{3t}I=2\int e^{3t}d(3t)\\e^{3t}I=2 e^{3t}+C\\I=2+Ce^{-3t}\]
8.1 Improper Integrals: Infinite Limits of Integration
One Infinite Limit and
Integrals with infinite limits of integration are improper integrals.
Both Limits Infinite
