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# 5.7: Integration Techniques

Created by: CK-12

## Integration by Substitution

There are a couple of tricky substitutions that are not intuitive. Here are some examples:

$\int \sqrt{2 - \sqrt{x}dx}$

The normal course of action is to make the expression inside of the radical equal to our new variable. This is the correct way to start but students may halt when they see the result:

$u = 2 - \sqrt{x} \ \ du = \frac{1}{2 \sqrt{x}}dx$

Which they will see as being useless to substitute back into the original integral. The trick here is to solve for x before taking the derivative:

$\sqrt{x} = 2 - u \rightarrow x = (2 - u) ^ 2 \rightarrow dx = -2 (2 - u) du$

Now we use the equation for $x$ and $dx$ to substitute back into our original integral:

$\int \sqrt {2 - \sqrt{(2 -u) ^ 2}}(- 4 + 2u)du = \int \sqrt{u} ( -4 + 2u)du = \int -4u ^{\frac{1}{2}} + 2u ^ {\frac{3}{2}}du= \frac{-2}{\sqrt{u}} + 3\sqrt{u} + c$

And finally substituting $x$ back in:

$\frac{-2}{\sqrt{2 - \sqrt{x}}} + 3 \sqrt{2 - \sqrt{x}} + c$

Another problem where we can apply the same “trick” is the, at first, innocent looking problem:

$\int \frac{x ^ 2 + 4}{x + 2}dx$

Normally the rule of thumb is to make the denominator equal to $u$, but in this case, that will not allow us to substitute out all of the $x$ variables in the problem. To do so, we will need to again get $x$ in terms of $u$.

$x + 2 = u && x = u -2 && dx = du$

$\int \frac{ (u -2) ^ 2 + 4}{u}du & = \int \frac{u ^ 2 - 4u + 8}{u} du = \int u - 4 + \frac{8}{u}du = \frac{1}{2} u ^ 2 -4u + 8 \ \text {ln} \ u + c \\& - \frac{1}{2} ( x + 2) ^ 2 - 4 (x + 2) + 8 \text (x + 2) + c$

## Integration by Parts

Typically the average person's experience has income arriving in discreet groupings, for example bi-monthly or monthly paychecks from employers. It is not the same for larger businesses, which owing to their size and the amount of their transactions think of income coming more as a stream. Businesses will often model the income with a function to help in making future projections. Since the income is often deposited into interest earning accounts, the value of a company can't be strictly computed just by how much money they are taking in currently. Economists will look at Present and Future values to determine the value of investments considering the “Time Preference” of money being worth more in-hand today than the same amount in the future. The Present and Future Values functions for businesses with income streams are:

Present Value:$V_p = \int\limits_{0}^{T} S (t) e ^ {-rt} dt$ Future Value: V_f = $\int\limits_{0} ^ {T} S(t) e ^ {r(T - t)}dt$

Where $S(t)$ is the income stream as a function of time, $T$ is the number of time periods (months, years) of evaluation and $r$ is the interest rate.

Find the present and future values for a seasonal sporting goods manufacturer who's income stream is modeled by the function $S(t) = -250 \cos(\frac{\pi}{6}t) + 625$, where $t$ is expressed in months and $S(t)$ is in thousands of dollars. The interest earned $.35 \%$ every month, and the term of the projection is $5$ years.

This problem is presented as an application that requires parts to solve. Many times the income stream is expressed as a constant or linear function, which may not require parts, but the multiplication of the exponential is frequently going to. I'll compute the Present Value here:

$V_p = \int\limits_{0}^{60}(-250 \cos \left ( \frac{\pi}{6} \right ) + 625)e ^ {-0.035t}dt = \int\limits_{0}^{60}-250 \cos \left ( \frac{\pi}{6} t \right ) e ^ {-0.035t} + 625 e ^ {-0.035t} dt$

The integral can be split, and the first term will require parts with

$u = e ^ {-.0035t} dv = \cos \left ( \frac{\pi}{6} t \right ) dt \rightarrow du = -.0035 e ^ {-.0035t}dt v = \frac{6}{\pi} \sin \left ( \frac{\pi}{6} t \right ) \\- 250 \int \limits_{0} ^ {60} \cos \left ( \frac{\pi}{6} t \right ) e ^ {-.0035t} dt = \frac{6}{\pi} \sin \left ( \frac{\pi}{6} t \right ) e ^ {-.0035t} + .0035 \frac{6}{\pi} \int \limits_ {0} ^ {60} \sin \left ( \frac{\pi}{6} t \right ) e ^ {-.0035t}dt$

Parts again with:

$u = e ^ {0.035t} dv = \sin \left ( \frac{\pi}{6} t \right )dt \rightarrow du = - .0035t e ^ {.0035t} dt v = - \frac{6}{\pi} \cos \left ( \frac{\pi}{6} t \right ) \\- 250 \int\limits_{0}^{60} \cos \left ( \frac{\pi}{6} t \right ) e ^ {- .0035t} dt \\= \frac{6}{\pi} \sin \left ( \frac{\pi}{6} t \right ) e ^ {- .0035t} + .0035 \frac{6}{\pi} \left ( \frac{- 6}{\pi} \cos \left ( \frac{\pi}{6} t \right ) e ^ { -.0035t} - .0035 \frac{6}{\pi} \int\limits_{0}^{60}\cos \left ( \frac{\pi}{6} t \right ) e ^ { - .0035t} dt \right )$

Distributing the numbers removes the parenthesis and allows us to “wrap around” the integral:

$.0035 \frac{-36}{\pi ^ 2} \cos \left ( \frac{\pi}{6} t \right ) e ^ {-.0035t} - .00004468 \int \limits_{0} ^ {60}\cos \left ( \frac{\pi}{6} t \right ) e ^ {-.0035t} dt \\- 249.999955 \int \limits_ {0} ^ {60} \cos \left ( \frac{\pi}{6} t \right ) e ^ {-.0035t} dt = \frac{6}{\pi} \sin \left ( \frac{\pi}{6} t \right ) e ^ {-.0035t} + .0035 \frac{-36}{\pi ^ 2} \cos \left ( \frac{\pi}{6} t \right ) e ^ {-.0035t}$

Finishing up:

$V_ p = \frac{250}{249.999955} ( \frac{6}{\pi} \sin \left ( \frac{\pi}{6} t \right ) e ^ {-.0035t} + .0035 \frac{-36}{\pi ^ 2} \cos \left ( \frac{\pi}{6} t \right ) e ^ {- .0035t} \big |_ {0} ^ {60} \approx 33823.60$

Very number intensive. The key here is record keeping, but the math is the same as simpler parts problems. The future value function works very much the same way.

## Integration by Partial Fractions

What if there is an irreducible quadratic term in the denominator after factoring? For example:

$\int \frac{10x + 2}{x ^ 3 - 5x ^ 2 + x - 5} dx = \int \frac{10x + 2}{(x ^ 2 + 1)(x - 5)} dx$

This is still a partial fractions problem. If there is an irreducible factor that is a quadratic in the denominator, then the numerator needs to be a linear term. In this case, the separation by partial fractions looks like:

$\frac{10x + 2}{(x ^ 2 + 1)(x - 5)} = \frac{Ax + b}{x ^ 2 + 1} + \frac{c}{x -5}$

Once the problem is set up correctly, it is solved in the same manner as all other partial fractions problems.

$\frac{10x + 2}{(x ^ 2 + 1)(x - 5)} = \frac{(Ax + B)(x - 5)}{(x ^ 2 + 1)(x - 5)} + \frac{C(x ^ 2 + 1)}{(x ^ 2 + 1)( x - 5)}$

After finding common denominators set the numerators equal

$10x + 2 = Ax ^ 2 + Bx - 5Ax - 5B + Cx ^ 2 + C$

Gather and factor terms with variables with the same power

$0x^2+10x+2=(A+C) x^2+(-5A+B)x+(-5B+C)$

Then set the coefficients of each variable equal on both sides of the equation

$0=A+C && 10=-5A+B && 2=-5B+C$

When solving $3$ variable systems and above, I nearly always use matrices on the calculator, as I make fewer mistakes than I do with substitution. Typical mistakes with substitution are going to be centered around distributing coefficients, especially negatives, correctly. After finding the value of each variable, plug those numbers back into original separation and integrate.

$\int \frac{-2x}{x ^ 2 + 1} + \frac{2}{x - 5} dx = - \text{ln} ( x ^ 2 + 1) + 2 \text {ln} ( x - 5) + c$

It should be pointed out that a major place of confusion for students is in the difference between $\frac{A}{(x + 1)} ^ 2$ and $\frac{Bx + c}{x ^ 2 + 1}$. The former is a repeated linear factor and the latter is a irreducible quadratic factor, and they must be treated differently. The technique listed above can also be extended for larger degree irreducible factors.

## Trigonometric Integrals

There are a couple of ways to solve a particular integral which will illustrate good practices with trig identities and integration.

Solve $\int \sin(x) \cos(x)dx$ three different ways.

It is possible that students can brainstorm the different ways, but it is also a good activity to assign different groups the methods of solution.

Method 1: Substitution

This is the most straightforward method.

$u = \sin(x) \rightarrow du = \cos(x) dx \\\int udu = \frac{1}{2} u ^ 2 + c = \frac{1}{2} \sin ^ 2 (x) + c$

Method 2: Integration by Parts

Since these are two functions that are multiplied, it makes sense to use parts:

$u & = \sin(x) && du = \cos(x)dx && dv = \cos(x)dx && v = \sin(x)\\& \int \sin(x) \cos(x)dx = \sin ^ 2 (x) - \int \sin(x)\cos(x) dx \\& 2 \int \sin(x)\cos(x) dx = \sin ^ 2 (x) + c\\& \int \sin(x)\cos(x) dx = \frac{1}{2} \sin ^ 2 (x) + c$

Method 3: Trig identities

The identities needed here are the double angle identities:$\sin(2x) = 2 \sin(x)\cos(x) , \cos(2x) = \cos ^ 2(x) - \sin ^ 2 (x)$ .

$& \int \sin(x)\cos(x) dx & = \frac{1}{2} \int \sin(2x) dx \\& = \frac{- 1}{4} \cos(2x) + c \\& = \frac{- 1}{4} ( \cos ^ 2 (x) - \sin ^ 2 (x)) + c\\& = \frac{-1}{4} ( 1 - \sin ^ 2 (x) - \sin ^ 2 (x))+ c \\& = \frac{-1}{4} ( 1 - 2 \sin ^ 2(x)) + c\\& = \frac{-1}{4} + \frac{1}{2} \sin(x) + c\\& = \frac{1}{2} \sin(x) + c\\$

The last example illustrates the importance of always including the constant added term, and remembering that any constant can be rolled into it, since it is not determined.

## Trig Substitution

The best problems in mathematics are often the ones that can be solved using different methods. There is something that captures my imagination about the truth and totality of the major theorems, like those presented by Euclid, that can be proven by straightedge and compass, and then thousands of years later with Galois groups. This is not nearly on the level of such classical problems, but it is valuable and entertaining for students to have the opportunity to verify facts using different methods. Especially those methods that may seem like they were dreamed up for the entertainment of torturing math students.

Here, we will examine the integral: $\int \frac{1}{1 - x ^ 2} dx$. If you ask the class without prompting, some may believe it looks like many of the problems they have just been working and that they should use a trig substitution. They would be correct. Others may recognize that the denominator can be factored as a difference of two squares, which allows the fraction to be separated using partial fractions. They are also correct. The class should show that the two methods give the same solution. This can be done either by asking every student to choose their preferred method, grouping students to work together with their preferred method, assigning a method or having everyone to work both methods on their own.

Partial Fractions:

$\frac{1}{(1 + x)(1 - x)} = \frac{A}{1 + x} + \frac{B}{1 - x} \rightarrow 1 = A - Ax + B + Bx \rightarrow 0 = -A + B && 1 = A + B \\\frac{1}{2} \int \frac{1}{1 + x} dx + \frac{1}{2} \int \frac{1}{1 - x} dx= \frac{1}{2} \text {ln} ( 1 + x) - \frac{1}{2} \text {ln} ( 1 -x ) + c$

Trig Sub:

This does not fit the substitution for sin exactly, but the subtraction indicates that the sine substitution is the one we need.

$x = \sin \phi && dx = \cos \phi d \phi \rightarrow \int \frac{\cos \phi}{1 - \sin ^ 2 \phi}d \phi = \int \frac{\cos \phi}{\cos ^ 2 \phi} d \phi = \int \sec \phi d \phi$

The unique method for taking this integral is outlined in the previous chapter's example $5$:

$\int \sec \phi d \phi = \int \frac{\sec ^ 2 \phi + \sec \phi \tan \phi}{\sec \phi + \tan \phi} d \phi \\u = \sec \phi + \tan \phi && du = \sec ^ 2 \phi + \sec \phi \tan \phi \\\int \frac{1}{u}du = \text {ln} u + c = \text {ln} ( \sec\phi + \tan\phi) + c = \text {ln} \left ( \frac{1}{\cos \phi} + \frac{\sin \phi}{\cos \phi} \right ) + c = \text {ln} \left ( \frac{1 + \sin \phi}{\cos \phi} \right ) + c$

Using the rules of logs, then substituting back in x using trig identities, we can find the same answer as above:

$\text {ln} ( 1 + \sin \phi) - \text{ln} ( \cos \phi) + c = \text {ln} ( 1 + x ) - \text {ln} ((1 - x ^ 2 ) ^ {\frac{1}{2}} ) + c \\& = \text {ln} ( 1 + x ) - \frac{1}{2} \text {ln} ((1 + x)(1 - x)) + c \\& = \text {ln} ( 1 + x) - \frac{1}{2} \text {ln} ( 1 + x) - \frac{1}{2}\text {ln} ( 1 -x ) + c\\& = \frac{1}{2} \text {ln} ( 1 + x) - \frac{1}{2} \text {ln} ( 1 - x) + c$

## Improper Integrals

Coulomb's Law is an equation that gives the electrostatic force between two charged particles. The scalar form of Coulomb's Law is:

$F = k \frac{q_1 q _2}{r ^ 2}$

where Coulomb's constant, $k = 8.9876 \times 10 ^ 9 \frac{Nm ^ 2}{C ^ 2}, q_1,q_2$ are the individual magnitudes of the two charges and $r$ is the distance between the two charges. This can be used to describe the force of attraction between a proton and an electron. In chemistry, ions have the ability to “take” electrons away from atoms. We can ask here, How much energy does it take to strip an electron from a hydrogen atom?

Reference sources state that the charge of both a proton and an electron is $1.6 \times 10 ^ {-19} C$, and the distance can be assumed to be the Bohr radius: $5.3 \times 10 ^ {-11} m$. Astute students may recognize that Coulomb's Law provides the force between the charges, not the energy required to move them, which would be expressed in joules, or the force times the distance traveled. Now, as the charges are spread apart, that also affects Coulomb's Law, as the force will get weaker, therefore it is not a simple multiplication. Hopefully students will recognize that in order to find a quantity as a product of a changing function over an interval they will need to integrate. One last problem, how far away do we need to take this electron to “strip” it away? To be safe, let's take it infinitely far away from the proton. Now the integral looks like this:

$E = \int\limits_{5.3 \times 10 ^ {-11}}^{\infty}(8.9876 \times 10 ^ 9 ) \frac{(1.6 \times 10 ^ {-19}) ^ 2}{r ^ 2} dr$

I placed all of the quantities in, but it will probably be easier to integrate using constant variables rather than using all of the very large, or very small numbers involved. The only thing to be careful of is to remember what is a constant, and what is the variable. Also, notice that this is an improper integral, so we will need to express it as a limit:

$E = k q_1q_2 \lim_{n \to \infty} \int\limits_{B}^{n} \frac{1}{r ^ 2} dr = k q_1q_2 \lim_{n \to \infty} \left ( \frac{-1}{n} - \frac{-1}{B} \right )$

Now we can see that as n approaches infinity, that term goes to zero, so the integral does converge. Substituting in the quantities left out:

$E = \frac{(8.9876 \times 10 ^ 9)(1.6 \times 10 ^ {-19}) ^ 2}{5.3 \times 10 ^ {-11}} \approx 4.34 \times 10 ^ {-8}$

We do expect the integral to converge. As the distance between the particles advances to infinity, the force becomes minimal, and with the squared term in the denominator, this is a classic converging integral.

## Ordinary Differential Equations

A common application of differential equation is fluid mixing problems. Given information of about the rate of increase or decrease of both the concentration and the fluid being mixed in sets up as a fairly common separable equation. Example:

A pond near a cement plant has been found to have a concentration hexavalent chromium ($(Cr VI)$ of $.72 \text {ppm}$. The volume of the pond is $1.17 \times 10 ^ 9 m ^ 3$, and there is a creek that carries contaminated water out that flows at a rate of $3$ cubic meters per second. Assuming fresh water with no contaminants is replaced in the pond, and all contaminants mix completely, how long will it take for the pond to return to the EPA specified limit of .1ppm of $Cr VI$?

We need to find the rate at which the chromium leaves the lake. Since the amount that leaves at any single time will depend on the current concentration, the rate that the contaminant leaves will be equal to the rate of water leaving times the concentration of the contaminant. Or put in variables:

$\frac{dC}{dt} = \frac{-rC}{V}$

Which is a separable differential equation:

$\frac{1}{C} dC = \frac{-r}{V}dt \rightarrow \text {ln} C = \frac{-r}{V} t + C_0 \rightarrow C = C_0 e ^ {\frac{-rt}{v}}$

This will allow us to calculate the time needed after putting in the initial condition $C_0$. To deal with more realistic time units, convert the flow rate of the creek to $7776000$ cubic meters per month. The since we can consider the current reading to be time zero, the initial conditions are:

$.72 = C_0 e ^ {\frac{-7776000 \times 0}{1.17 \times 10 ^ 9}}\rightarrow .72 = C_0$

Solving then for the time:

$.1 = .72 e ^ {\frac{-7776000 \times 0}{1.17 \times 10 ^ 9}}\rightarrow .1389 = e ^ {-.0065t} \rightarrow \text {ln}(.1389) = - .0065t \rightarrow 303.69 = t$

Therefore the pond will be back down to safe levels in just over $303$ months, or $25$ years.

## Date Created:

Feb 23, 2012

Apr 29, 2014
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